NCERT Solutions Class 11 Maths Chapter 9

Q.1 Write the first five terms of the sequence whose nth terms is:
a_n = n(n+2)

Ans

$\begin{array}{l}\text{We have}\\ {\text{a}}_{\text{n}}=\text{n}(\text{n}+\text{2})\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}=\text{1}(\text{1}+\text{2})\\ =3\\ {\text{a}}_{\text{2}}=\text{2}(\text{2}+\text{2})\\ =8\\ {\text{a}}_{\text{3}}=\text{3}(\text{3}+\text{2})\\ =15\\ {\text{a}}_{\text{4}}=\text{4}(\text{4}+\text{2})\\ =24\\ {\text{a}}_{\text{5}}=\text{5}(\text{5}+\text{2})\\ =35\\ \mathrm{Therefore},\text{the first five terms are: 3, 8, 15, 24, 35.}\end{array}$

Q.2

$\begin{array}{l}\mathbf{\text{Write the first five terms of the sequence whose nth terms is:}}\\ {\mathbf{\text{a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{\mathbf{\text{n}}}{\mathbf{\text{n+1}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{We have}\\ {\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+1}\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}=\frac{1}{1+1}\\ =\frac{1}{2}\\ {\text{a}}_{\text{2}}=\frac{2}{2+1}\\ =\frac{2}{3}\\ {\text{a}}_{\text{3}}=\frac{3}{3+1}\\ =\frac{3}{4}\\ {\text{a}}_{\text{4}}=\frac{4}{4+1}\\ =\frac{4}{5}\\ {\text{a}}_{\text{5}}=\frac{5}{5+1}\\ =\frac{5}{6}\\ \mathrm{Therefore},\text{the first five terms are:}\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}.\end{array}$

Q.3 Write the first five terms of the sequence whose nth terms is:
a_n = 2ⁿ

Ans

${\text{We have, a}}_{\text{n}}={\text{2}}^{\text{n}}$ $\begin{array}{l}\mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}={\text{2}}^{\text{1}}\\ =2\\ {\text{a}}_{\text{2}}={\text{2}}^{\text{2}}\\ =4\\ {\text{a}}_{\text{3}}={\text{2}}^{\text{3}}\\ =8\\ {\text{a}}_{\text{4}}={\text{2}}^{\text{4}}\\ =16\\ {\text{a}}_{\text{5}}={\text{2}}^{\text{5}}\\ =32\\ \mathrm{Therefore},\text{the first five terms are: 2, 4, 8, 16, 32.}\end{array}$

Q.4

$\begin{array}{l}\mathbf{\text{Write the first five terms of the sequence whose nth terms is:}}\\ {\mathbf{\text{a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{\mathbf{\text{2n}}\mathbf{-}\mathbf{\text{3}}}{\mathbf{\text{6}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{We have}\\ {\mathrm{a}}_{\mathrm{n}}=\frac{2\mathrm{n}-3}{6}\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}=\frac{2\left(1\right)-3}{6}\\ =-\frac{1}{6}\\ {\text{a}}_{\text{2}}=\frac{2\left(2\right)-3}{6}\\ =\frac{1}{6}\end{array}$ $\begin{array}{l}{\text{a}}_{\text{3}}=\frac{2\left(3\right)-3}{6}\\ =\frac{3}{6}\\ =\frac{1}{2}\\ {\text{a}}_{\text{4}}=\frac{2\left(4\right)-3}{6}\\ =\frac{5}{6}\\ {\text{a}}_{\text{5}}=\frac{2\left(5\right)-3}{6}\\ =\frac{7}{6}\\ \mathrm{Therefore},\text{the first five terms are:}-\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}.\end{array}$

Q.5 Write the first five terms of the sequence whose nth terms is a_n = (– 1)^n–1 5ⁿ⁺¹.

Ans

$\begin{array}{l}{\text{We have a}}_{\text{n}}={(-\text{1})}^{\text{n}-\text{1}}{\text{5}}^{\text{n}+\text{1}}\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}={(–\text{1})}^{\text{1}-\text{1}}{\text{5}}^{\text{1}+\text{1}}\\ =25\\ {\text{a}}_{\text{2}}={(–\text{1})}^{\text{2}-\text{1}}{\text{5}}^{\text{2}+\text{1}}\\ =-125\\ {\text{a}}_{\text{3}}={(-\text{1})}^{\text{3}-\text{1}}{\text{5}}^{\text{3}+\text{1}}\\ =625\end{array}$ $\begin{array}{l}{\text{a}}_{\text{4}}={(-\text{1})}^{\text{4}-\text{1}}{\text{5}}^{\text{4}+\text{1}}\\ =-3125\\ {\text{a}}_{\text{5}}={(-\text{1})}^{\text{5}-\text{1}}{\text{5}}^{\text{5}+\text{1}}\\ =15625\\ \mathrm{Therefore},\text{the first five terms are:25,}-\text{125,625,}-3125,15625.\end{array}$

Q.6

$\begin{array}{l}\mathbf{\text{Write the first five terms of the sequence whose nth terms is}}\\ {\mathbf{\text{a}}}_{\mathbf{\text{n}}}\mathbf{\text{=n}}\frac{{\mathbf{\text{n}}}^{\mathbf{\text{2}}}\mathbf{\text{+5}}}{\mathbf{\text{4}}}\end{array}$

Ans

$\begin{array}{l}\text{We have}{\mathrm{a}}_{\mathrm{n}}=\mathrm{n}\frac{{\mathrm{n}}^2+5}{4}\\ \text{Substituting n = 1, 2, 3, 4, 5 respectively, we get}\\ {\text{a}}_{\text{1}}=\left(1\right)\frac{{\left(1\right)}^2+5}{4}\\ =\frac{6}{4}=\frac{3}{2}\\ {\text{a}}_{\text{2}}=\left(2\right)\frac{{\left(2\right)}^2+5}{4}\\ =\frac{9}{2}\\ {\text{a}}_{\text{3}}=\left(3\right)\frac{{\left(3\right)}^2+5}{4}\\ =\frac{42}{4}=\frac{21}{2}\\ {\text{a}}_{\text{4}}=\left(4\right)\frac{{\left(4\right)}^2+5}{4}\\ =21\\ {\text{a}}_{\text{5}}=\left(5\right)\frac{{\left(5\right)}^2+5}{4}\\ =\frac{150}{4}\\ =\frac{75}{2}\\ \text{Therefore, the first five terms are:}\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}.\end{array}$

Q.7

$\begin{array}{l}\mathbf{\text{Find the indicated terms in each of the sequences given below}}\\ {\mathbf{\text{whose n}}}^{\mathbf{\text{th}}}\mathbf{\text{terms are:}}\\ \mathbf{1}\mathbf{.}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{4}\mathbf{n}\mathbf{–}\mathbf{3}\mathbf{;}{\mathbf{a}}_{\mathbf{17}}\mathbf{,}{\mathbf{a}}_{\mathbf{24}}\\ {\mathbf{\text{2. a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{{\mathbf{\text{n}}}^{\mathbf{\text{2}}}}{{\mathbf{\text{2}}}^{\mathbf{\text{n}}}}{\mathbf{\text{;\hspace{0.17em}\hspace{0.17em}a}}}_{\mathbf{\text{7}}}\\ \mathbf{3}\mathbf{.}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}{(–1)}^{\mathbf{n}\mathbf{–}\mathbf{1}}{\mathbf{n}}^{\mathbf{3}}\mathbf{;}{\mathbf{a}}_{\mathbf{9}}\\ {\mathbf{\text{4. a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{\mathbf{\text{n}}(\text{n}-\text{2})}{\mathbf{\text{n+3}}}{\mathbf{\text{; a}}}_{\mathbf{\text{20}}}\end{array}$

Ans

1.

$\begin{array}{l}\text{Given:}\\ {\text{a}}_{\text{n}}=\text{4n}–\text{3}\\ \mathrm{Substituting}\text{n=17 and 24 respectively, we get}\\ {\text{a}}_{\text{17}}=\text{4}\left(17\right)-\text{3}\\ =\text{68}-\text{3}\\ =\text{65}\\ {\text{a}}_{\text{24}}=\text{4}\left(24\right)-\text{3}\\ =96-3\\ =93\end{array}$

2.

$\begin{array}{l}\text{Given:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2}{2^{\mathrm{n}}}\\ \mathrm{Substituting}\mathrm{n}=7\text{, we get}\\ {\text{a}}_{\text{7}}=\frac{7^2}{2^7}\\ =\frac{49}{128}\\ {\text{Thus, the value of a}}_{\text{7}}\text{is}\frac{49}{128}\text{.}\end{array}$

3.

$\begin{array}{l}{\text{Given:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{n}}={(-\text{1})}^{\text{n}–\text{1}}{\text{n}}^{\text{3}}\\ \mathrm{Substituting}\mathrm{n}=9\text{, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{9}}={(-\text{1})}^{\text{9}–\text{1}}{\left(9\right)}^{\text{3}}\\ ={(-\text{1})}^8\left(729\right)\\ =729\\ \mathrm{Thus},{\text{the value of a}}_{\text{9}}\text{is 729.}\end{array}$ \begin{array}{l}\text{4.}\end{array} $\begin{array}{l}\text{Given:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3}\\ \mathrm{Substituting}\mathrm{n}=20\text{, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}a}}_{\text{20}}=\frac{20(20-2)}{20+3}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{20\left(18\right)}{23}\\ =\frac{360}{23}\\ \mathrm{Thus},{\text{the value of a}}_{\text{20}}\text{is}\frac{360}{23}\text{.}\end{array}$

Q.8

$\begin{array}{l}\mathbf{\text{Write the first five terms of each of the sequences given below and}}\\ \mathbf{\text{obtain the corresponding series:}}\\ {\mathbf{\text{1. a}}}_{\mathbf{\text{1}}}\mathbf{\text{=}}\mathbf{3}{\mathbf{\text{, a}}}_{\mathbf{\text{n}}}{\mathbf{\text{=3a}}}_{\mathbf{\text{n-1}}}\mathbf{\text{+2, for all n}}\mathbf{>}\mathbf{1}\\ {\mathbf{\text{2. a}}}_{\mathbf{\text{1}}}\mathbf{\text{=}}\mathbf{–}{\mathbf{\text{1, a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{{\mathbf{\text{a}}}_{\mathbf{\text{n-1}}}}{\mathbf{\text{n}}}\mathbf{\text{, n}}\mathbf{\ge }\mathbf{\text{2}}\\ \mathbf{3}\mathbf{.}{\mathbf{a}}_{\mathbf{1}}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}{{\mathbf{a}}_{\mathbf{n}}}_{\mathbf{–}\mathbf{1}}\mathbf{–}\mathbf{1}\mathbf{,}\mathbf{n}\mathbf{>}\mathbf{2}\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:{\mathrm{a}}_{\mathrm{n}}=3{\mathrm{a}}_{\mathrm{n}}-1+2{\text{and a}}_{\text{1}}=\text{3}\\ \mathrm{Substituting}\text{n}=\text{2, we get}\\ {\mathrm{a}}_2=3{{\mathrm{a}}_2}_{-1}+2\\ =3{\mathrm{a}}_1+2\\ =3\left(3\right)+2[\mathrm{Putting}{\text{a}}_{\text{1}}=\text{3}]\\ =11\\ \mathrm{Substituting}\text{n}=\text{3, we get}\\ {\mathrm{a}}_3=3{{\mathrm{a}}_3}_{-1}+2\\ =3{\mathrm{a}}_2+2\\ =3\left(11\right)+2[\mathrm{Putting}{\text{a}}_{\text{2}}=\text{11}]\\ =35\\ \mathrm{Substituting}\text{n}=\text{4, we get}\\ {\mathrm{a}}_4=3{{\mathrm{a}}_4}_{-1}+2\\ =3{\mathrm{a}}_3+2\\ =3\left(35\right)+2[\mathrm{Putting}{\text{a}}_{\text{3}}=\text{35}]\\ =107\end{array}$ $\begin{array}{l}\mathrm{Substituting}\text{n}=\text{5, we get}\\ {\mathrm{a}}_5=3{{\mathrm{a}}_5}_{-1}+2\\ =3{\mathrm{a}}_4+2\\ =3\left(107\right)+2[\mathrm{Putting}{\text{a}}_{\text{4}}=\text{107}]\\ =323\\ \mathrm{Thus},\text{the first five terms are 3, 11, 35, 107 and 323.}\\ \text{Corresponding series is:\hspace{0.17em} 3}+\text{11}+\text{35}+\text{107}+\text{323}+...\end{array}$

2.

$\begin{array}{l}\mathrm{Given}:{\mathrm{a}}_1=-1,{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}–1}}{\mathrm{n}},\mathrm{n}\ge 2\\ \mathrm{Substituting}\text{n}=\text{2, we get}\\ {\mathrm{a}}_2=\frac{{\mathrm{a}}_{2–1}}{2}\\ =\frac{{\mathrm{a}}_1}{2}\\ =\frac{-1}{2}[\mathrm{Putting}{\text{a}}_{\text{1}}=-\text{1}]\\ \mathrm{Substituting}\text{n}=\text{3, we get}\\ {\mathrm{a}}_3=\frac{{\mathrm{a}}_{3–1}}{3}\\ =\frac{{\mathrm{a}}_2}{3}\\ =\frac{\left(\frac{-1}{2}\right)}{3}[\mathrm{Putting}{\text{a}}_{\text{2}}=\frac{-1}{2}]\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{-1}{6}\\ \mathrm{Substituting}\text{n}=\text{4, we get}\\ {\mathrm{a}}_4=\frac{{\mathrm{a}}_{4–1}}{4}\\ =\frac{{\mathrm{a}}_3}{4}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{\left(\frac{-1}{6}\right)}{4}[\mathrm{Putting}{\text{a}}_{\text{3}}=\frac{-1}{6}]\\ =\frac{-1}{24}\\ \mathrm{Substituting}\text{n}=\text{5, we get}\\ {\mathrm{a}}_5=\frac{{\mathrm{a}}_{5–1}}{5}\\ =\frac{{\mathrm{a}}_4}{5}\\ =\frac{\left(\frac{-1}{24}\right)}{5}[\mathrm{Putting}{\text{a}}_{\text{4}}=\frac{-1}{24}]\\ =\frac{-1}{120}\\ \mathrm{Thus},\text{the first five terms are}-\text{1,}\frac{-\text{1}}{2}\text{,}\frac{-\text{1}}{6}\text{,}\frac{-\text{1}}{24}\text{and}\frac{-\text{1}}{120}\text{.}\\ \text{The corresponding series is:}\\ (-\text{1})+\left(\frac{-\text{1}}{2}\right)+\left(\frac{-\text{1}}{6}\right)+\left(\frac{-\text{1}}{24}\right)+\left(\frac{-\text{1}}{120}\right)+...\end{array}$

$\begin{array}{l}\mathrm{Given}:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{n}}={{\text{a}}_{\text{n}}}_{–\text{1}}-\text{1},{\mathrm{a}}_1={\mathrm{a}}_2=2,\text{n}>\text{2}\\ \mathrm{Substituting}\text{n}=\text{3, we get}\\ {\mathrm{a}}_3={{\text{a}}_{\text{3}}}_{–\text{1}}-\text{1}\\ ={\text{a}}_{\text{2}}-\text{1}\\ =2-1[\mathrm{Putting}{\text{a}}_{\text{2}}=2]\\ =1\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{Substituting}\text{n}=\text{4, we get}\\ {\mathrm{a}}_4={{\text{a}}_{\text{4}}}_{–\text{1}}-\text{1}\\ ={\text{a}}_{\text{3}}-\text{1}\\ =1-1[\mathrm{Putting}{\text{a}}_{\text{3}}=1]\\ =0\\ \mathrm{Substituting}\text{n}=\text{5, we get}\\ {\mathrm{a}}_5={{\text{a}}_{\text{4}}}_{–\text{1}}-\text{1}\\ ={\text{a}}_{\text{3}}-\text{1}\\ =0-1[\mathrm{Putting}{\text{a}}_{\text{4}}=0]\\ =-1\\ \mathrm{Thus},\text{the first five terms of series are:}\\ \text{2,\hspace{0.17em}2,\hspace{0.17em}1,\hspace{0.17em}0,}-\text{1}\\ \text{The series is:\hspace{0.17em}\hspace{0.17em}2}+\text{\hspace{0.17em}2}+\text{\hspace{0.17em}1}+\text{\hspace{0.17em}0}+(-\text{1})+...\end{array}$

Q.9

$\begin{array}{l}{\mathbf{\text{The Fibonacci sequence is defined by 1=a}}}_{\mathbf{\text{1}}}{\mathbf{\text{=a}}}_{\mathbf{\text{2}}}{\mathbf{\text{and a}}}_{\mathbf{\text{n}}}{\mathbf{\text{=a}}}_{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}{\mathbf{\text{+a}}}_{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{2}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}n>2.}}\\ \mathbf{\text{Find}}\frac{{\mathbf{\text{a}}}_{\mathbf{\text{n+1}}}}{{\mathbf{\text{a}}}_{\mathbf{\text{n}}}}\mathbf{\text{, for n=1,2,3,4,5.}}\end{array}$

Ans

$\begin{array}{l}\text{TheFibonaccisequenceisdefinedby}\\ {\text{1=a}}_{\text{1}}{\text{=a}}_{\text{2}}{\text{anda}}_{\text{n}}{\text{=a}}_{\text{n}-\text{1}}{\text{+a}}_{\text{n}-\text{2}}\text{,n>2.}\\ \text{Substitutingn=3,weget}\\ {\mathrm{a}}_3={\mathrm{a}}_2+{\mathrm{a}}_1\\ =1+1\\ =2\\ \text{Forn=1,weget}\\ \frac{{\text{a}}_{\text{1+1}}}{{\text{a}}_{\text{1}}}\text{=}\frac{{\text{a}}_{\text{2}}}{{\text{a}}_{\text{1}}}\\ \text{=}\frac{\text{1}}{\text{1}}\\ \frac{{\text{a}}_{\text{2}}}{{\text{a}}_{\text{1}}}\text{=1}\\ \text{and}\frac{{\mathrm{a}}_3}{{\mathrm{a}}_2}=\frac{2}{1}=2\\ \frac{{\text{a}}_{\text{n+1}}}{{\text{a}}_{\text{n}}}\text{=}\frac{{\text{a}}_{\text{n}}{\text{+a}}_{\text{n}-\text{1}}}{{\text{a}}_{\text{n-1}}{\text{+a}}_{\text{n}-\text{2}}}\\ \text{Substituting}\mathrm{n}=3\text{,weget}\\ \frac{{\text{a}}_{\text{3+1}}}{{\text{a}}_{\text{3}}}\text{=}\frac{{\text{a}}_{\text{3}}{\text{+a}}_{\text{3}-\text{1}}}{{\text{a}}_{\text{3}-\text{1}}{\text{+a}}_{\text{3}-\text{2}}}\\ \frac{{\text{a}}_{\text{4}}}{{\text{a}}_{\text{3}}}\text{=}\frac{{\text{a}}_{\text{3}}{\text{+a}}_{\text{2}}}{{\text{a}}_{\text{2}}{\text{+a}}_{\text{1}}}\\ \text{=}\frac{\left({\text{a}}_{\text{2}}{\text{+a}}_{\text{1}}\right){\text{+a}}_{\text{2}}}{{\text{a}}_{\text{2}}{\text{+a}}_{\text{1}}}\left[{\mathrm{a}}_3={\mathrm{a}}_2+{\mathrm{a}}_1\right]\\ \text{=}\frac{\left(\text{1+1}\right)\text{+1}}{\text{1+1}}\\ \text{=}\frac{\text{3}}{\text{2}}\\ \mathrm{Substituting}\text{n}=4,\text{we get}\\ {\text{a}}_{\text{4}}={\text{a}}_3+{\text{a}}_2\\ =2+1\\ =3\\ \frac{{\text{a}}_{4+1}}{{\text{a}}_4}=\frac{{\text{a}}_4+{\text{a}}_3}{{\text{a}}_3+{\text{a}}_2}\\ \frac{{\text{a}}_5}{{\text{a}}_4}=\frac{3+2}{2+1}\\ =\frac{5}{3}\\ \mathrm{Substituting}\text{n}=5,\text{weget}\\ {\text{a}}_{\text{5}}={\text{a}}_4+{\text{a}}_3\\ =3+2\\ =5\\ \frac{{\text{a}}_{5+1}}{{\text{a}}_5}=\frac{{\text{a}}_5+{\text{a}}_4}{{\text{a}}_4+{\text{a}}_3}\\ \frac{{\text{a}}_6}{{\text{a}}_5}=\frac{5+3}{3+2}\\ =\frac{8}{5}\\ \mathrm{Thus},\text{therequiredtermsare:}1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5}.\end{array}$

Q.10 Find the sum of odd integers from 1 to 2001.

Ans

$\begin{array}{l}\mathrm{The}\text{odd integers from 1 to 2001 are:}\\ 1,3,5,7,9,11,13,...,1999,2001\\ \mathrm{a}=1,\mathrm{d}=3-1=2\\ \mathrm{l}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 2001=1+(\mathrm{n}-1)2\\ 2001-1=(\mathrm{n}-1)2\\ \mathrm{n}-1=\frac{2000}{2}\\ \mathrm{n}=1000+1\\ =1001\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})\\ {\mathrm{S}}_{1001}=\frac{1001}{2}(1+2001)\\ {\mathrm{S}}_{1001}=\frac{1001}{2}\left(2002\right)\\ =1001\left(1001\right)\\ =1002001\\ \mathrm{Thus},\text{the sum of odd integers from 1 to 2001 is 1002001.}\end{array}$

Q.11 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Ans

$\begin{array}{l}\mathrm{The}\text{multiples of 5 from 100 to 1000 are:}\\ 105,105,110,115,120,125,...,995\\ \mathrm{a}=100,\mathrm{d}=105-100=5\\ \mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ 995=105+\left(\mathrm{n}-1\right)5\\ 995-105=\left(\mathrm{n}-1\right)5\\ \mathrm{n}-1=\frac{890}{5}\\ \mathrm{n}=178+1\\ =179\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ {\mathrm{S}}_{179}=\frac{179}{2}\left(105+995\right)\\ =\frac{179}{2}\times \left(1100\right)\\ =98450\\ \mathrm{Thus},\text{the sum of the natural numbers which are multiple of 5}\\ \text{between 100 to 1000 is}98450.\end{array}$

Q.12 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20^th term is –112.

Ans

$\begin{array}{l}\mathrm{First}\text{term of A.P.}=\text{2}\\ \text{Let common difference}=\mathrm{d}\\ \mathrm{According}\text{to given condition:}\\ \text{Sum of first 5 terms}=\frac{1}{4}\left(\mathrm{Sum}\text{of next 5 terms}\right)\\ {\mathrm{S}}_5=\frac{1}{4}({\mathrm{S}}_{10}-{\mathrm{S}}_5)\\ \Rightarrow 4{\mathrm{S}}_5={\mathrm{S}}_{10}-{\mathrm{S}}_5\\ \Rightarrow 5{\mathrm{S}}_5={\mathrm{S}}_{10}\end{array}$ $\begin{array}{l}\Rightarrow 5\left[\frac{5}{2}\{2\times 2+(5-1)\mathrm{d}\}\right]=\left[\frac{10}{2}\{2\times 2+(10-1)\mathrm{d}\}\right]\\ [âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}]\\ \Rightarrow \frac{25}{2}(4+4\mathrm{d})=5(4+9\mathrm{d})\\ \Rightarrow 25(2+2\mathrm{d})=5(4+9\mathrm{d})\\ \Rightarrow 50+50\mathrm{d}=20+45\mathrm{d}\\ \Rightarrow 5\mathrm{d}=20-50\\ \Rightarrow \mathrm{d}=-\frac{30}{5}\\ =-6\\ \mathrm{Now},\text{sum of 20 terms.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{20}}=2+(20-1)\times -6[âˆ\mu {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}]\\ =2-114\\ =-112\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term is}-\text{112.}\end{array}$

Q.13

$\begin{array}{l}\mathbf{\text{How many terms of the A.P.\hspace{0.17em}\hspace{0.17em}}}\mathbf{-}\mathbf{\text{6,}}\mathbf{-}\frac{\mathbf{\text{11}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{-}\mathbf{\text{5,}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em} are needed to give}}\\ \mathbf{\text{the sum}}\mathbf{-}\mathbf{\text{25?}}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given A.P. is}-6,-\frac{11}{2},-5,...\\ \mathrm{Let}\text{the sum of n terms of A.P. is}-\text{25.Then,}\end{array}$ $\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}\\ \mathrm{a}=-6\\ \mathrm{d}=-\frac{11}{2}-(-6)\\ =-\frac{11}{2}+6=\frac{1}{2}\\ -25=\frac{\mathrm{n}}{2}\{2\times -6+(\mathrm{n}-1)\times \frac{1}{2}\}[âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}]\\ -25=\frac{\mathrm{n}}{2}\{-12+(\mathrm{n}-1)\times \frac{1}{2}\}\\ -100=\mathrm{n}(-24+\mathrm{n}-1)\\ -100=-25\mathrm{n}+{\mathrm{n}}^2\\ {\mathrm{n}}^2-25\mathrm{n}+100=0\\ (\mathrm{n}-20)(\mathrm{n}-5)=0\\ \Rightarrow \mathrm{n}=20,5\\ \mathrm{Thus},\text{the number of terms of A.P. is 5 or 20.}\end{array}$

Q.14

$\begin{array}{l}{\mathbf{\text{In an A.P., if p}}}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{p}}}{\mathbf{\text{and q}}}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{q}}}\mathbf{\text{, prove that the sum of first}}\\ \mathbf{\text{pq terms is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\left(\text{pq+1}\right)\mathbf{\text{, where p}}\mathbf{\ne }\mathbf{\text{q.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P.}=\mathrm{a}\\ \mathrm{Let}\text{common difference of A.P.}=\mathrm{d}\\ \mathrm{Since},{\text{T}}_{\text{p}}=\frac{1}{\mathrm{q}}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{a}+(\mathrm{p}-1)\mathrm{d}=\frac{1}{\mathrm{q}}...\left(\mathrm{i}\right)\\ \mathrm{And},{\text{T}}_{\text{q}}=\frac{1}{\mathrm{p}}\\ \Rightarrow \mathrm{a}+(\mathrm{q}-1)\mathrm{d}=\frac{1}{\mathrm{p}}...\left(\mathrm{ii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ (\mathrm{p}-1-\mathrm{q}+1)\mathrm{d}=\frac{1}{\mathrm{q}}-\frac{1}{\mathrm{p}}\\ \Rightarrow (\mathrm{p}-\mathrm{q})\mathrm{d}=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{pq}}\\ \Rightarrow \mathrm{d}=\frac{1}{\mathrm{pq}}\\ \mathrm{Substituting}\text{value of d in equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{a}+(\mathrm{p}-1)\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ \Rightarrow \mathrm{a}+\frac{\mathrm{p}}{\mathrm{pq}}-\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ \Rightarrow \mathrm{a}=\frac{1}{\mathrm{pq}}\\ \mathrm{Now},{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\{2\times \frac{1}{\mathrm{pq}}+(\mathrm{pq}-1)\frac{1}{\mathrm{pq}}\}\\ [âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}]\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\{2\times \frac{1}{\mathrm{pq}}+(\mathrm{pq}-1)\frac{1}{\mathrm{pq}}\}\\ =\frac{\mathrm{pq}}{2}(\frac{2}{\mathrm{pq}}+1-\frac{1}{\mathrm{pq}})\\ =\frac{\mathrm{pq}}{2}(\frac{1}{\mathrm{pq}}+1)\\ =\frac{1}{2}(1+\mathrm{pq})\\ \mathrm{Thus},\text{the sum of pq terms is}\frac{1}{2}(\mathrm{pq}+1).\end{array}$

Q.15 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Ans

$\begin{array}{l}\begin{array}{l}\mathrm{Given}\text{A.P. 25},\text{22},\text{19},\dots \\ \mathrm{First}\text{term of A.P.}=\text{25}\\ \text{Common difference of A.P.}=22-25\end{array}\\ \begin{array}{l}=-3\\ \mathrm{Let}\text{sum of n terms of A.P.}=116\end{array}\\ \text{i}.\text{e}.,{\text{S}}_{\text{n}}=116\\ \frac{\text{n}}{2}\left\{2\times 25+\left(\text{n}-1\right)\left(-3\right)\right\}=116\\ \Rightarrow \text{n}\left(50-3\text{n}+3\right)=232\\ \begin{array}{l}\Rightarrow \left(53\text{n}-3{\text{n}}^2\right)=232\\ \Rightarrow 3{\text{n}}^2-53\text{n}+232=0\end{array}\\ \Rightarrow 3{\text{n}}^2-24\text{n}-29\text{n}+232=0\\ \Rightarrow 3\text{n}\left(\text{n}-8\right)-29\left(\text{n}-8\right)=0\\ \begin{array}{l}\Rightarrow \left(\text{n}-8\right)\left(3\text{n}-29\right)=0\\ \Rightarrow \text{n}=8,\frac{29}{3}\end{array}\\ \mathrm{Since},\text{n is a natural number.So,}\\ \begin{array}{l}\text{n}=\text{8}\\ \text{Last term of A.P.}=\text{a}+\left(\text{n}-1\right)\text{d}\end{array}\\ =25+\left(8-1\right)\left(-3\right)\\ =25-21\\ =4\\ \mathrm{Thus},\text{the last term of A.P. is 4.}\end{array}$

Q.16 Find the sum to n terms of the A.P., whose k^th term is 5k + 1.

Ans

$\begin{array}{l}{\mathrm{k}}^{\mathrm{th}}\text{term of A.P.}=\text{5k+1}\\ \Rightarrow {\mathrm{T}}_{\mathrm{k}}=\text{5k+1}\\ \mathrm{Substituting}\text{k}=1,2,3,...\text{respectively.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{1}}=5\left(1\right)+1\\ =6\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{2}}=5\left(2\right)+1\\ =11\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{3}}=5\left(3\right)+1\\ =16\\ \mathrm{Common}\text{difference}={\mathrm{T}}_2-{\mathrm{T}}_1\\ =11-6\\ =5\end{array}$ $\begin{array}{l}\mathrm{Sum}{\text{of n terms,S}}_{\text{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}\\ =\frac{\mathrm{n}}{2}\{2\times 6+(\mathrm{n}-1)5\}\\ =\frac{\mathrm{n}}{2}(12+5\mathrm{n}-5)\\ =\frac{\mathrm{n}}{2}(7+5\mathrm{n})\\ =\frac{\mathrm{n}}{2}(5\mathrm{n}+7)\\ \mathrm{Thus},\text{the sum of n terms is}\frac{\mathrm{n}}{2}(5\mathrm{n}+7).\end{array}$

Q.17 If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference.

Ans

$\begin{array}{l}\mathrm{Given},{\mathrm{S}}_{\mathrm{n}}=(\mathrm{pn}+{\mathrm{qn}}^2)\\ \mathrm{Putting}\text{n}=1,2,3\mathrm{respectively},\mathrm{we}\mathrm{get}\\ {\mathrm{S}}_1=(\mathrm{p}\times 1+\mathrm{q}\times 1^2)\\ =\mathrm{p}+\mathrm{q}\\ {\mathrm{S}}_2=(\mathrm{p}\times 2+\mathrm{q}\times 2^2)\\ =2\mathrm{p}+4\mathrm{q}\\ {\mathrm{S}}_3=(\mathrm{p}\times 3+\mathrm{q}\times 3^2)\\ =3\mathrm{p}+9\mathrm{q}\\ {\mathrm{T}}_1={\mathrm{S}}_2-{\mathrm{S}}_1\\ =(2\mathrm{p}+4\mathrm{q})-(\mathrm{p}+\mathrm{q})\end{array}$ $\begin{array}{l}=\mathrm{p}+3\mathrm{q}\\ {\mathrm{T}}_2={\mathrm{S}}_3-{\mathrm{S}}_2\\ =(3\mathrm{p}+9\mathrm{q})-(2\mathrm{p}+4\mathrm{q})\\ =\mathrm{p}+5\mathrm{q}\\ \mathrm{Common}\text{difference,}\\ \text{d}={\mathrm{T}}_2-{\mathrm{T}}_1\\ =(\mathrm{p}+5\mathrm{q})-(\mathrm{p}+3\mathrm{q})\\ =2\mathrm{q}\\ \mathrm{Thus},\text{common difference is 2q.}\end{array}$

Q.18 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18^th terms.

Ans

$\begin{array}{l}\mathrm{Since},\\ \frac{\mathrm{Sum}\text{of n terms of an A.P.}}{\mathrm{Sum}\text{of n terms of another A.P.}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ \Rightarrow \frac{\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}}{\frac{\mathrm{n}}{2}\{2\mathrm{A}+(\mathrm{n}-1)\mathrm{D}\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ \Rightarrow \frac{\{\mathrm{a}+\left(\frac{\mathrm{n}-1}{2}\right)\mathrm{d}\}}{\{\mathrm{A}+\left(\frac{\mathrm{n}-1}{2}\right)\mathrm{D}\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}...\left(\mathrm{i}\right)\\ \mathrm{Since},{\text{T}}_{\text{18}}=\text{a}+\text{17d. So, Substituting}\frac{\mathrm{n}-1}{2}=17\end{array}$ $\begin{array}{l}\mathrm{or}\mathrm{n}=35\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{\{\mathrm{a}+17\mathrm{d}\}}{\{\mathrm{A}+17\mathrm{D}\}}=\frac{\text{5}\times \text{35}+\text{4}}{\text{9}\times \text{35}+\text{6}}\\ \frac{{\text{18}}^{\text{th}}\text{term of an A.P.}}{{\text{18}}^{\text{th}}\text{term of another A.P.}}=\frac{\text{175}+\text{4}}{\text{315}+\text{6}}\\ =\frac{179}{321}\\ \mathrm{Thus},{\text{the ratio of 18}}^{\text{th}}\text{terms of two A.P. is 179:321.}\end{array}$

Q.19 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Ans

$\begin{array}{l}\text{first term of A.P.}=\text{a}\\ \mathrm{and}\text{common difference of A.P.}=\text{d}\\ {\mathrm{S}}_{\mathrm{p}}={\mathrm{S}}_{\mathrm{q}}\left(\mathrm{Given}\right)\\ \Rightarrow \frac{\mathrm{p}}{2}\{2\mathrm{a}+(\mathrm{p}-1)\mathrm{d}\}=\frac{\mathrm{q}}{2}\{2\mathrm{a}+(\mathrm{q}-1)\mathrm{d}\}\\ \Rightarrow 2\mathrm{ap}+\mathrm{p}(\mathrm{p}-1)\mathrm{d}=2\mathrm{aq}+\mathrm{q}(\mathrm{q}-1)\mathrm{d}\\ \Rightarrow 2\mathrm{ap}-2\mathrm{aq}=\mathrm{q}(\mathrm{q}-1)\mathrm{d}-\mathrm{p}(\mathrm{p}-1)\mathrm{d}\\ \Rightarrow 2\mathrm{a}(\mathrm{p}-\mathrm{q})=({\mathrm{q}}^2-\mathrm{q}-{\mathrm{p}}^2+\mathrm{p})\mathrm{d}\\ \Rightarrow 2\mathrm{a}(\mathrm{p}-\mathrm{q})=\{\mathrm{p}-\mathrm{q}-({\mathrm{p}}^2-{\mathrm{q}}^2)\mathrm{d}\}\\ \Rightarrow 2\mathrm{a}(\mathrm{p}-\mathrm{q})=\{(\mathrm{p}-\mathrm{q})-(\mathrm{p}-\mathrm{q})(\mathrm{p}+\mathrm{q})\mathrm{d}\}\\ \Rightarrow 2\mathrm{a}(\mathrm{p}-\mathrm{q})=(\mathrm{p}-\mathrm{q})\{1-(\mathrm{p}+\mathrm{q})\mathrm{d}\}\\ \Rightarrow 2\mathrm{a}=\{1-(\mathrm{p}+\mathrm{q})\mathrm{d}\}\end{array}$ $\begin{array}{l}\mathrm{Now},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\{2\mathrm{a}+(\mathrm{p}+\mathrm{q}-1)\mathrm{d}\}\\ {\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}[\{1-(\mathrm{p}+\mathrm{q})\}\mathrm{d}+(\mathrm{p}+\mathrm{q}-1)\mathrm{d}]\\ {\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\{1-(\mathrm{p}+\mathrm{q})+(\mathrm{p}+\mathrm{q})-1\}\mathrm{d}\\ \Rightarrow {\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=0\\ \mathrm{Thus},\text{the sum of}(\mathrm{p}+\mathrm{q})\text{terms is zero.}\end{array}$

Q.20

$\begin{array}{l}\text{Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.}\\ \text{Prove that}\frac{\text{a}}{\text{p}}\left(\text{q}-\text{r}\right)\text{+}\frac{\text{b}}{\text{q}}\left(\text{r}-\text{p}\right)\text{+}\frac{\text{c}}{\text{r}}\left(\text{p}-\text{q}\right)\text{=0.}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be A and common difference be d.}\\ \mathrm{Given}:{\mathrm{S}}_{\mathrm{p}}=\mathrm{a}\Rightarrow \mathrm{a}=\frac{\mathrm{p}}{2}\{2\mathrm{A}+(\mathrm{p}-1)\mathrm{d}\}\\ \Rightarrow \frac{\mathrm{a}}{\mathrm{p}}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{p}-1)\mathrm{d}\}\\ {\mathrm{S}}_{\mathrm{q}}=\mathrm{b}\Rightarrow \mathrm{b}=\frac{\mathrm{p}}{2}\{2\mathrm{A}+(\mathrm{q}-1)\mathrm{d}\}\\ \Rightarrow \frac{\mathrm{b}}{\mathrm{q}}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{q}-1)\mathrm{d}\}\\ \text{and}{\mathrm{S}}_{\mathrm{r}}=\mathrm{c}\Rightarrow \mathrm{c}=\frac{\mathrm{r}}{2}\{2\mathrm{A}+(\mathrm{r}-1)\mathrm{d}\}\\ \Rightarrow \frac{\mathrm{c}}{\mathrm{r}}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{r}-1)\mathrm{d}\}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}}{\mathrm{p}}(\mathrm{q}-\mathrm{r})+\frac{\mathrm{b}}{\mathrm{q}}(\mathrm{r}-\mathrm{p})+\frac{\mathrm{c}}{\mathrm{r}}(\mathrm{p}-\mathrm{q})\\ =\frac{1}{2}\{2\mathrm{A}+(\mathrm{p}-1)\mathrm{d}\}(\mathrm{q}-\mathrm{r})+\frac{1}{2}\{2\mathrm{A}+(\mathrm{q}-1)\mathrm{d}\}(\mathrm{r}-\mathrm{p})\\ +\frac{1}{2}\{2\mathrm{A}+(\mathrm{r}-1)\mathrm{d}\}(\mathrm{p}-\mathrm{q})\\ =\frac{1}{2}\left[\begin{array}{l}2\mathrm{A}(\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q})+\\ \mathrm{d}\{(\mathrm{p}-1)(\mathrm{q}-\mathrm{r})+(\mathrm{q}-1)(\mathrm{r}-\mathrm{p})+(\mathrm{r}-1)(\mathrm{p}-\mathrm{q})\}\end{array}\right]\\ =\frac{1}{2}\{2\mathrm{A}\left(0\right)+\mathrm{d}\left(\begin{array}{l}\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{qp}-\mathrm{r}+\mathrm{p}\\ +\mathrm{rp}-\mathrm{rq}-\mathrm{p}+\mathrm{q}\end{array}\right)\}\\ =\frac{1}{2}(0+0)\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.21 The ratio of the sums of m and n terms of an A.P. is m² : n². Show that the ratio of m^th and n^th term is (2m – 1) : (2n – 1).

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of an A.P.}=\text{a}\\ \text{and commong difference of A.P.}=\mathrm{d}\\ \mathrm{It}\text{is given that:}\\ {\text{S}}_{\text{m}}{\text{:S}}_{\text{n}}={\mathrm{m}}^2:{\mathrm{n}}^2\\ \Rightarrow \frac{{\mathrm{S}}_{\mathrm{m}}}{{\mathrm{S}}_{\mathrm{n}}}=\frac{{\mathrm{m}}^2}{{\mathrm{n}}^2}\end{array}$ $\begin{array}{l}\Rightarrow \frac{\frac{\mathrm{m}}{2}\{2\mathrm{a}+(\mathrm{m}-1)\mathrm{d}\}}{\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}}=\frac{{\mathrm{m}}^2}{{\mathrm{n}}^2}\\ \Rightarrow \frac{\{2\mathrm{a}+(\mathrm{m}-1)\mathrm{d}\}}{\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}}=\frac{\mathrm{m}}{\mathrm{n}}\\ \Rightarrow 2\mathrm{an}+(\mathrm{mn}-\mathrm{n})\mathrm{d}=2\mathrm{am}+(\mathrm{mn}-\mathrm{m})\mathrm{d}\\ \Rightarrow 2\mathrm{an}-2\mathrm{am}=(\mathrm{mn}-\mathrm{m})\mathrm{d}-(\mathrm{mn}-\mathrm{n})\mathrm{d}\\ \Rightarrow 2\mathrm{a}(\mathrm{n}-\mathrm{m})=\mathrm{d}(\mathrm{mn}-\mathrm{m}-\mathrm{mn}+\mathrm{n})\\ \Rightarrow 2\mathrm{a}(\mathrm{n}-\mathrm{m})=\mathrm{d}(-\mathrm{m}+\mathrm{n})\\ \Rightarrow 2\mathrm{a}=\mathrm{d}\\ \mathrm{Now},\\ \frac{{\mathrm{T}}_{\mathrm{m}}}{{\mathrm{T}}_{\mathrm{n}}}=\frac{\mathrm{a}+(\mathrm{m}-1)2\mathrm{a}}{\mathrm{a}+(\mathrm{n}-1)2\mathrm{a}}\\ =\frac{\mathrm{a}+(\mathrm{m}-1)2\mathrm{a}}{\mathrm{a}+(\mathrm{n}-1)2\mathrm{a}}\\ =\frac{1+2\mathrm{m}-2}{1+2\mathrm{n}-2}\\ =\frac{2\mathrm{m}-1}{2\mathrm{n}-1}\\ \mathrm{Thus},{\text{the ratio of m}}^{\text{th}}{\text{and n}}^{\text{th}}\text{term is}(2\mathrm{m}-1):(2\mathrm{n}-1).\end{array}$

Q.22 If the sum of n terms of an A.P. is 3n² + 5n and its m^th term is 164, find the value of m.

Ans

$\begin{array}{l}\mathrm{It}\text{is given that}\\ {\text{S}}_{\text{n}}={\text{3n}}^{\text{2}}+\text{5n}\\ \text{Replace n by n}-\text{1, we get}\\ {\text{S}}_{\text{n-1}}=\text{3}{\left(\text{n}-\text{1}\right)}^{\text{2}}+\text{5}\left(\text{n}-1\right)\\ ={\text{3n}}^{\text{2}}-6\text{n}+\text{3}+\text{5n}-5\\ ={\text{3n}}^{\text{2}}-\text{n}-2\\ \mathrm{So},{\text{T}}_{\text{n}}={\text{S}}_{\text{n}}-{\text{S}}_{\text{n-1}}\\ =\left({\text{3n}}^{\text{2}}+\text{5n}\right)-\left({\text{3n}}^{\text{2}}-\text{n}-2\right)\\ =6\text{n}+2\\ \mathrm{Since},{\text{T}}_{\text{m}}=164\\ \mathrm{So},\text{6m}+\text{2}=\text{164}\\ \Rightarrow \text{m}=\frac{164-2}{6}\\ =\frac{162}{6}\\ =27\\ \mathrm{Thus},\text{the value of m is 27.}\end{array}$

Q.23 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Ans

$\begin{array}{l}\mathrm{Let}{\text{5 numbers between 8 and 26 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{and A}}_{\text{5}}\text{.}\\ {\text{Then, A.P. is 8, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{, A}}_{\text{5}}\text{,\hspace{0.17em}26.}\\ âˆ\mu \mathrm{Last}\text{term}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 26=8+(7-1)\mathrm{d}\end{array}$ $\begin{array}{l}26=8+6\mathrm{d}\\ \mathrm{d}=\frac{26-8}{6}\\ =\frac{18}{6}\\ =3\\ \mathrm{Then},{\mathrm{A}}_1=8+\mathrm{d}\\ =8+3\\ =11\\ {\mathrm{A}}_2=8+2\mathrm{d}\\ =8+2\left(3\right)\\ =14\\ {\mathrm{A}}_3=8+3\mathrm{d}\\ =8+3\left(3\right)\\ =17\\ {\mathrm{A}}_4=8+4\mathrm{d}\\ =8+4\left(3\right)\\ =20\end{array}$ \begin{array}{l}\mathrm{}\end{array} $\begin{array}{l}{\mathrm{A}}_5=8+5\mathrm{d}\\ =8+5\left(3\right)\\ =23\\ \mathrm{Thus},\text{the 5 terms between 8 and 26 are 11, 14, 17, 20 and 23.}\end{array}$

Q.24

$\mathbf{\text{If}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}is the A.M. between a and b, then find the value of n.}}$

Ans

$\begin{array}{l}\mathrm{A}.\mathrm{M}.\text{of a and b}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \Rightarrow \left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}\right)=2\left({\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}\right)\\ \Rightarrow {\mathrm{a}}^{\mathrm{n}}+{\mathrm{ab}}^{\mathrm{n}-1}+{\mathrm{ba}}^{\mathrm{n}-1}+{\mathrm{b}}^{\mathrm{n}}=2{\mathrm{a}}^{\mathrm{n}}+2{\mathrm{b}}^{\mathrm{n}}\\ \Rightarrow {\mathrm{ab}}^{\mathrm{n}-1}-{\mathrm{b}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}-{\mathrm{ba}}^{\mathrm{n}-1}\\ \Rightarrow {\mathrm{b}}^{\mathrm{n}-1}\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^{\mathrm{n}-1}\left(\mathrm{a}-\mathrm{b}\right)\\ \Rightarrow {\mathrm{b}}^{\mathrm{n}-1}={\mathrm{a}}^{\mathrm{n}-1}\\ \Rightarrow {\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{n}-1}=1={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^0\\ \Rightarrow \mathrm{n}=1\end{array}$

Q.25 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7^th and (m – 1)^th numbers is 5 : 9. Find the value of m.

Ans

$\begin{array}{l}\mathrm{Let}{\text{m numbers between 1 and 31 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{.}\\ {\text{Then, A.P. is 1, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{,\hspace{0.17em}\hspace{0.17em}31.}\\ âˆ\mu \mathrm{Last}\text{term}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 31=1+(\mathrm{m}+2-1)\mathrm{d}\\ 30=(\mathrm{m}+1)\mathrm{d}\\ \mathrm{d}=\frac{30}{\mathrm{m}+1}\\ \mathrm{Then},{\mathrm{A}}_1=1+\mathrm{d}\\ =1+\frac{30}{\mathrm{m}+1}\\ =\frac{\mathrm{m}+31}{\mathrm{m}+1}\\ {\mathrm{A}}_7=1+7\mathrm{d}\end{array}$ $\begin{array}{l}=1+7\left(\frac{30}{\mathrm{m}+1}\right)\\ =\frac{\mathrm{m}+211}{\mathrm{m}+1}\\ {\mathrm{A}}_{\mathrm{m}-1}=1+(\mathrm{m}-1)\mathrm{d}\\ =1+(\mathrm{m}-1)\left(\frac{30}{\mathrm{m}+1}\right)\\ =\frac{\mathrm{m}+1+30(\mathrm{m}-1)}{\mathrm{m}+1}\\ =\frac{31\mathrm{m}-29}{\mathrm{m}+1}\\ \mathrm{According}\text{to question:}\\ \frac{{\mathrm{A}}_7}{{\mathrm{A}}_{\mathrm{m}-1}}=\frac{\left(\frac{\mathrm{m}+211}{\mathrm{m}+1}\right)}{\left(\frac{31\mathrm{m}-29}{\mathrm{m}+1}\right)}\\ \frac{5}{9}=\frac{\mathrm{m}+211}{31\mathrm{m}-29}\\ 155\mathrm{m}-145=9\mathrm{m}+1899\\ 155\mathrm{m}-9\mathrm{m}=1899+145\\ 146\mathrm{m}=2044\\ \mathrm{m}=\frac{2044}{146}\\ =14\\ \mathrm{Thus},\text{the value of m is 14.}\end{array}$

Q.26 The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

Ans

$\begin{array}{l}\mathrm{The}\text{difference between any two consecutive}\\ \text{angles}=5\mathrm{°}\\ \mathrm{The}\text{smallest angle of a polygon}=120°\\ \mathrm{Let}\text{number of sides in given polygon be n.Then,}\\ \text{Sum of angles of a polygon}=(\mathrm{n}-2)180\mathrm{°}\\ \mathrm{The}\text{series formed by angles of polygon is:}\\ \text{120°,125°,130°,135°,140°,\hspace{0.17em}}...\end{array}$ $\begin{array}{l}\mathrm{Sum}\text{of n angles of polygon of n sides}=\frac{\mathrm{n}}{2}\{2\times 120\mathrm{°}+(\mathrm{n}-1)\times 5\mathrm{°}\}\\ =\frac{\mathrm{n}}{2}\{240\mathrm{°}+5\mathrm{°}\mathrm{n}-5\mathrm{°}\}\\ =\frac{\mathrm{n}}{2}\{235\mathrm{°}+5\mathrm{°}\mathrm{n}\}\\ \mathrm{Therefore},\\ \frac{\mathrm{n}}{2}\{235\mathrm{°}+5\mathrm{°}\mathrm{n}\}=(\mathrm{n}-2)180\mathrm{°}\\ \Rightarrow 5\mathrm{°}{\mathrm{n}}^2+235\mathrm{°}\mathrm{n}=360\mathrm{°}\mathrm{n}-720\mathrm{°}\\ \Rightarrow 5\mathrm{°}{\mathrm{n}}^2-125\mathrm{°}\mathrm{n}+720\mathrm{°}=0\\ \Rightarrow {\mathrm{n}}^2-25\mathrm{°}\mathrm{n}+144\mathrm{°}=0\\ \Rightarrow (\mathrm{n}-9)(\mathrm{n}-16)=0\\ \Rightarrow \mathrm{n}=9,16\\ \mathrm{Number}\text{of sides in polygon is 9 because number of angles in}\\ \text{a polygon is equal to the number of sides. Sum of 16 angles}\\ \text{will be equal to sum of 9 angles if some angles are 0° or}\\ \text{negative,}\mathrm{which}\text{is impossible. Therefore number of sides in}\\ \text{the polygon is 9.}\end{array}$

Q.27

${\mathbf{\text{Find the 20}}}^{\mathbf{\text{th}}}{\mathbf{\text{and n}}}^{\mathbf{\text{th}}}\mathbf{\text{terms of the G.P.}}\frac{\mathbf{\text{5}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\frac{\mathbf{\text{5}}}{\mathbf{\text{4}}}\mathbf{\text{,}}\frac{\mathbf{\text{5}}}{\mathbf{\text{8}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}$

Ans

The\text{given G}\text{.P}\text{. is:}\frac{5}{2},\frac{5}{4},\frac{5}{8},\mathrm{\dots } $\begin{array}{l}\mathrm{a}=\frac{5}{2}\\ \mathrm{r}=\frac{\left(\frac{5}{4}\right)}{\left(\frac{5}{2}\right)}=\frac{1}{2}\\ {\mathrm{T}}_{20}={\mathrm{ar}}^{20-1}[âˆ\mu {\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}]\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{19}\\ =\frac{5}{2^{20}}\\ \mathrm{and}{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{\mathrm{n}-1}\\ =\frac{5}{2^{\mathrm{n}}}\end{array}$

Q.28 Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

Ans

$\begin{array}{l}\mathrm{Common}\text{ratio of given G.P.}=2\\ 8^{\mathrm{th}}\text{term of A.P.}=192\\ {\mathrm{ar}}^{8-1}=192\\ \mathrm{a}.2^7=192\\ \mathrm{a}=\frac{192}{2^7}=1.5\\ {12}^{\mathrm{th}}\text{term of A.P.}={\mathrm{ar}}^{12-1}\end{array}$ $\begin{array}{l}=1.5{\left(2\right)}^{11}\\ =1.5\times 2048\\ =3072\end{array}$

Q.29 The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Ans

$\begin{array}{l}\text{Let first term of G.P. be a and common ratio be r.}\\ {\text{\hspace{0.17em}\hspace{0.17em}5}}^{\text{th}}\text{term of G.P.}=\mathrm{p}\\ \Rightarrow {\mathrm{T}}_5=\mathrm{p}\\ \Rightarrow \mathrm{p}={\mathrm{ar}}^4\\ {\text{\hspace{0.17em}\hspace{0.17em}8}}^{\text{th}}\text{term of G.P.}=\mathrm{q}\\ \Rightarrow {\mathrm{T}}_8=\mathrm{q}\\ \Rightarrow \mathrm{q}={\mathrm{ar}}^7\\ {\text{11}}^{\text{th}}\text{term of G.P.}=\mathrm{s}\\ \Rightarrow {\mathrm{T}}_{11}=\mathrm{s}\\ \Rightarrow \mathrm{s}={\mathrm{ar}}^{10}\\ \text{L.H.S.}={\mathrm{q}}^2\\ ={\left({\mathrm{ar}}^7\right)}^2\\ ={\mathrm{a}}^2{\mathrm{r}}^{14}\\ \text{R.H.S.=}\mathrm{ps}\\ =\left({\mathrm{ar}}^4\right)\left({\mathrm{ar}}^{10}\right)\\ ={\mathrm{a}}^2{\mathrm{r}}^{14}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}L.H.S.=R.H.S.}\\ \text{Hence proved.}\end{array}$

Q.30 The 4^th term of a G.P. is square of its second term, and the first term is –3. Determine its 7^th term.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=-3\\ \mathrm{Let}\text{common ratio of G.P.}=\mathrm{r}\\ \mathrm{According}\text{to given condition,}\\ {\mathrm{T}}_4={\left({\mathrm{T}}_2\right)}^2\\ \Rightarrow {\mathrm{ar}}^3={\left(\mathrm{ar}\right)}^2\\ \Rightarrow (-3){\mathrm{r}}^3={(-3)}^2{\mathrm{r}}^2\\ \Rightarrow \mathrm{r}=-3\\ \mathrm{Then},{\mathrm{T}}_7={\mathrm{ar}}^6\\ =(-3){(-3)}^6\\ ={(-3)}^7\\ =-2187\\ \mathrm{Thus},{\text{7}}^{\text{th}}\text{term of G.P. is}-2187.\end{array}$

Q.31

$\begin{array}{l}\mathbf{\text{Which term of the following sequences:}}\\ \left(\text{a}\right)\mathbf{\text{. 2, 2}}\sqrt{\mathbf{\text{2}}}\mathbf{\text{, 4,}}\mathbf{.}\mathbf{..}\mathbf{\text{is 128?}}\\ \left(\text{b}\right)\mathbf{\text{. 3, 3}}\sqrt{\mathbf{\text{3}}}\mathbf{\text{, 3,}}\mathbf{.}\mathbf{..}\mathbf{\text{is 729?}}\\ \left(\text{c}\right)\mathbf{\text{. \hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{3}}}\mathbf{\text{,\hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{9}}}\mathbf{\text{,\hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{27}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em} is \hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{19683}}}\mathbf{\text{?}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Let}{\text{t}}_{\text{n}}=128,\text{a}=2,\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} r}=\frac{2\sqrt{2}}{2}=\sqrt{2}\end{array}$ $\begin{array}{l}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ 128=2{\left(\sqrt{2}\right)}^{\mathrm{n}-1}\\ \Rightarrow 2^7={\left(2\right)}^{1+\frac{\mathrm{n}-1}{2}}\\ \Rightarrow 7=1+\frac{\mathrm{n}-1}{2}\\ \Rightarrow 6\times 2=\mathrm{n}-1\Rightarrow \mathrm{n}=13\\ \mathrm{Thus},{\text{128 is 13}}^{\text{th}}\text{term of G.P.}\\ \left(\mathrm{b}\right)\mathrm{Let}{\text{t}}_{\text{n}}=729,\text{a}=\sqrt{3},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} r}=\frac{3}{\sqrt{3}}=\sqrt{3}\\ {\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ 729=\sqrt{3}{\left(\sqrt{3}\right)}^{\mathrm{n}-1}\\ \Rightarrow 3^6={\left(3\right)}^{\frac{1}{2}+\frac{\mathrm{n}-1}{2}}\\ \Rightarrow 6=\frac{1}{2}+\frac{\mathrm{n}-1}{2}\\ \Rightarrow 6\times 2=1+\mathrm{n}-1\\ \Rightarrow \mathrm{n}=12\\ \mathrm{Thus},729{\text{is 12}}^{\text{th}}\text{term of G.P.}\\ \left(\mathrm{c}\right)\mathrm{Let}{\text{t}}_{\text{n}}=729,\text{a}=\frac{1}{3},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}=\frac{\left(\frac{1}{9}\right)}{\left(\frac{1}{3}\right)}=\frac{1}{3}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ \frac{1}{19683}=\frac{1}{3}{\left(\frac{1}{3}\right)}^{\mathrm{n}-1}\\ \Rightarrow {\left(\frac{1}{3}\right)}^9={\left(\frac{1}{3}\right)}^{\mathrm{n}}\\ \Rightarrow 9=\mathrm{n}\\ \Rightarrow \mathrm{n}=9\\ \mathrm{Thus},\frac{1}{19683}{\text{is 9}}^{\text{th}}\text{term of G.P.}\end{array}$

Q.32

$\mathbf{\text{For what values of x, the numbers\hspace{0.17em}\hspace{0.17em}}}\mathbf{-}\frac{\mathbf{\text{2}}}{\mathbf{\text{7}}}\mathbf{\text{,x,}}\mathbf{-}\frac{\mathbf{\text{7}}}{\mathbf{\text{2}}}\mathbf{\text{are in G.P.?}}$

Ans

$\begin{array}{l}-\frac{\text{2}}{\text{7}}\text{,x,}-\frac{\text{7}}{\text{2}}\text{are in G.P., then}\\ \text{Common ratio}=\frac{\mathrm{x}}{-\left(\frac{2}{7}\right)}\\ =-\frac{7\mathrm{x}}{2}\\ \text{Also, common ratio}=\frac{\left(-\frac{7}{2}\right)}{\mathrm{x}}\\ =-\frac{7}{2\mathrm{x}}\\ \text{Then,}\\ -\frac{7\mathrm{x}}{2}=-\frac{7}{2\mathrm{x}}\\ \Rightarrow {\mathrm{x}}^2=1\\ \Rightarrow \mathrm{x}=\pm 1\end{array}$

Q.33

$\begin{array}{l}\mathbf{\text{Find the sum to indicated number of terms in each of the geometric}}\\ \mathbf{\text{progressions given below.}}\\ \mathbf{\text{i. 0.15, 0.015, 0.0015,… 20 terms.}}\\ \mathbf{\text{ii.}}\sqrt{\mathbf{\text{7}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}}}\sqrt{\mathbf{\text{21}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}3}}\sqrt{\mathbf{\text{7}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}n\hspace{0.17em}terms.}}\\ \mathbf{\text{iii. 1,}}\mathbf{-}{\mathbf{\text{\hspace{0.17em}a, a}}}^{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{-}{\mathbf{\text{\hspace{0.17em}a}}}^{\mathbf{\text{3}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}\mathbf{\text{n terms (if a}}\mathbf{\ne }\mathbf{-}\mathbf{\text{1).}}\\ {\mathbf{\text{iv. x}}}^{\mathbf{\text{3}}}{\mathbf{\text{, x}}}^{\mathbf{\text{5}}}{\mathbf{\text{, x}}}^{\mathbf{\text{7}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}\mathbf{\text{n terms (if x}}\mathbf{\ne }\mathbf{\text{± 1).}}\end{array}$

Ans

$\begin{array}{l}\mathrm{i}.\\ 0.\text{15},0.0\text{15},0.00\text{15},\dots \text{2}0\text{terms}\\ \mathrm{Since},{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}(1-{\mathrm{r}}^{\mathrm{n}})}{1-\mathrm{r}}\\ \mathrm{a}=0.15,\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}r}=\frac{0.015}{0.15}=0.1\\ {\mathrm{S}}_{\mathrm{n}}=\frac{0.15\{1-{(0.1)}^{20}\}}{1-0.1}\\ =\frac{0.15\{1-{(0.1)}^{20}\}}{0.9}\\ =\frac{15}{90}\{1-{(0.1)}^{20}\}\\ =\frac{1}{6}\{1-{(0.1)}^{20}\}\end{array}$ $\begin{array}{l}\mathrm{i}\mathrm{i}.\\ \sqrt{7},\sqrt{21},3\sqrt{7},...\mathrm{n}\mathrm{terms}\\ \mathrm{Since},{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{a}=\sqrt{7},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} r}=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\sqrt{7}\{{\left(\sqrt{3}\right)}^{\mathrm{n}}-1\}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\\ =\frac{\sqrt{7}\{{\left(\sqrt{3}\right)}^{\mathrm{n}}-1\}}{3-1}\times (\sqrt{3}+1)\end{array}$ \begin{array}{l}\mathrm{}\end{array} $\begin{array}{l}\mathrm{iii}.\\ 1,-\mathrm{a},{\mathrm{a}}^2,-{\mathrm{a}}^3,...\mathrm{nterms}\\ \mathrm{First}\text{term of G.P.}=1\\ \mathrm{Common}\text{ratio of G.P.}=\frac{-\mathrm{a}}{1}\\ =-\mathrm{a}\\ \mathrm{Since},{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}(1-{\mathrm{r}}^{\mathrm{n}})}{1-\mathrm{r}}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{1\{1-{(-\mathrm{a})}^{\mathrm{n}}\}}{1+\mathrm{a}}\end{array}$ \begin{array}{l}\mathrm{}\end{array} $\begin{array}{l}\mathrm{iv}.\\ {\mathrm{x}}^3,{\mathrm{x}}^5,{\mathrm{x}}^7,...\mathrm{n}\mathrm{terms}\\ \mathrm{First}\text{term of G.P.}={\mathrm{x}}^3\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{Common}\text{ratio of G.P.}=\frac{{\mathrm{x}}^5}{{\mathrm{x}}^3}\\ ={\mathrm{x}}^2\\ \mathrm{Since},{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{x}}^3\{1-{\left({\mathrm{x}}^2\right)}^{\mathrm{n}}\}}{1-{\mathrm{x}}^2}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{x}}^3(1-{\mathrm{x}}^{2\mathrm{n}})}{1-{\mathrm{x}}^2}\end{array}$

Q.34

$\mathbf{Evaluate}\mathbf{:}\sum _{k=1}^{11}(2+3^k).$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{}\mathrm{have},\\ \sum _{\mathrm{k}=1}^{11}(2+3^{\mathrm{k}})=(2+3^1)+(2+3^2)+(2+3^3)+...+(2+3^{11})\\ =2\times 11+(3^1+3^2+3^3+...+3^{11})\\ =22+\frac{3(3^{11}-1)}{3-1}\\ =22+\frac{3}{2}(3^{11}-1)\end{array}$

Q.35

$\begin{array}{l}\mathbf{\text{The sum of first three terms of a G.P. is}}\frac{\mathbf{\text{39}}}{\mathbf{\text{10}}}\mathbf{\text{and their product is 1.}}\\ \mathbf{\text{Find the common ratio and the terms.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{three terms in A.P. are}\frac{\mathrm{a}}{\mathrm{r}},\mathrm{a},\mathrm{ar}.\\ \mathrm{Then}\text{according to given conditions:}\\ \frac{\text{a}}{\mathrm{r}}+\mathrm{a}+\mathrm{ar}=\frac{39}{10}\\ \mathrm{a}(\frac{1}{\mathrm{r}}+1+\mathrm{r})=\frac{39}{10}...\left(\mathrm{i}\right)\\ \mathrm{and}\frac{\mathrm{a}}{\mathrm{r}}\times \mathrm{a}\times \mathrm{ar}=1\\ {\mathrm{a}}^3=1\\ \Rightarrow \mathrm{a}=1\\ \mathrm{Substituting}\text{value of a in equation}\left(\mathrm{i}\right),\mathrm{we}\text{get}\\ 1(\frac{1}{\mathrm{r}}+1+\mathrm{r})=\frac{39}{10}\\ \Rightarrow \left(\frac{1+\mathrm{r}+{\mathrm{r}}^2}{\mathrm{r}}\right)=\frac{39}{10}\\ 10({\mathrm{r}}^2+\mathrm{r}+1)=39\mathrm{r}\\ 10{\mathrm{r}}^2+10\mathrm{r}+10=39\mathrm{r}\\ 10{\mathrm{r}}^2-29\mathrm{r}+10=0\\ 10{\mathrm{r}}^2-25\mathrm{r}-4\mathrm{r}+10=0\\ 5\mathrm{r}(2\mathrm{r}-5)-2(2\mathrm{r}-5)=0\\ (2\mathrm{r}-5)(2\mathrm{r}-5)=0\\ \Rightarrow \mathrm{r}=\frac{5}{2},\frac{5}{2}\end{array}$ $\begin{array}{l}\mathrm{Thus},\text{first term of G.P. is 1 and common difference is}\frac{5}{2}.\\ \mathrm{Three}\text{terms are:\hspace{0.17em}}\frac{1}{\frac{5}{2}},1,1\times \frac{5}{2}\Rightarrow \frac{2}{5},1,\frac{5}{2}\\ \mathrm{Or}\frac{1}{\frac{2}{5}},1,1\times \frac{2}{5}\Rightarrow \frac{5}{2},1,\frac{2}{5}\end{array}$

Q.36 How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

Ans

$\begin{array}{l}\text{G}.\text{P}.\text{is 3},{\text{3}}^{\text{2}},{\text{3}}^{\text{3}},\dots \\ \mathrm{a}=3,\text{r}=\frac{3^2}{3}=3{\text{and S}}_{\text{n}}=120\\ {\mathrm{S}}_{\mathrm{n}}=\mathrm{a}\frac{({\mathrm{r}}^{\mathrm{n}}-1)}{(\mathrm{r}-1)}\end{array}$ $\begin{array}{l}120=3\left(\frac{3^{\mathrm{n}}-1}{3-1}\right)\\ 120=\frac{3}{2}(3^{\mathrm{n}}-1)\\ 120\times \frac{2}{3}=(3^{\mathrm{n}}-1)\\ 81=3^{\mathrm{n}}\\ \Rightarrow 3^4=3^{\mathrm{n}}\\ \Rightarrow \mathrm{n}=4\\ \mathrm{Thus},\text{4 terms of G.P. are needed to give the sum 120.}\end{array}$

Q.37 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.
Determine the first term, the common ratio and the sum to n terms of the G.P.

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of G.P. is a and common ratio is r. Then,}\\ {\text{six terms of G.P. are: a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}{\text{, ar}}^{\text{5}}{\text{, ar}}^{\text{6}}\text{.}\\ \text{According to first condition:}\\ \text{a}+\text{ar}+{\text{ar}}^{\text{2}}=16\\ \Rightarrow \mathrm{a}(1+\mathrm{r}+{\mathrm{r}}^2)=16...\left(\mathrm{i}\right)\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}}^{\text{3}}+{\text{ar}}^{\text{4}}+{\text{ar}}^{\text{5}}=128\\ \Rightarrow {\mathrm{ar}}^3(1+\mathrm{r}+{\mathrm{r}}^2)=128...\left(\mathrm{ii}\right)\\ \mathrm{Dividing}\text{equation}\left(\mathrm{ii}\right)\text{by equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{{\mathrm{ar}}^3(1+\mathrm{r}+{\mathrm{r}}^2)}{\mathrm{a}(1+\mathrm{r}+{\mathrm{r}}^2)}=\frac{128}{16}\\ \Rightarrow {\mathrm{r}}^3=8\\ \Rightarrow \mathrm{r}=2\\ \mathrm{Putting}\text{r}=\text{2 in equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{a}(1+2+2^2)=16\\ \mathrm{a}=\frac{16}{7}\text{and r}=2\\ \mathrm{Sum}\text{of n terms is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{(\mathrm{r}-1)}\end{array}$ $\begin{array}{l}=\frac{\frac{16}{7}(2^{\mathrm{n}}-1)}{(2-1)}\\ =\frac{16}{7}(2^{\mathrm{n}}-1)\end{array}$

Q.38 Given a G.P. with a = 729 and 7^th term 64, determine S₇.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\left(\text{a}\right)\\ =729\\ 7^{\mathrm{th}}\mathrm{term}\text{of G.P.}={\mathrm{T}}_7\\ =64\\ \mathrm{Then},{\text{\hspace{0.17em}ar}}^{\text{6}}=64\\ \Rightarrow 729{\mathrm{r}}^6=64\\ {\mathrm{r}}^6=\frac{64}{729}\\ {\mathrm{r}}^6=\frac{2^6}{3^6}\\ \Rightarrow \mathrm{r}=\frac{2}{3}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}(1-{\mathrm{r}}^{\mathrm{n}})}{(1-\mathrm{r})}\\ \Rightarrow {\mathrm{S}}_7=\frac{729(1-\frac{2^{\mathrm{n}}}{3^{\mathrm{n}}})}{(1-\frac{2}{3})}\end{array}$ $\begin{array}{l}\Rightarrow {\mathrm{S}}_7=729\times 3(1-\frac{2^7}{3^7})\\ \Rightarrow {\mathrm{S}}_7=729\times 3(1-\frac{128}{2187})\\ \Rightarrow {\mathrm{S}}_7=2187\left(\frac{2187-128}{2187}\right)\\ \Rightarrow {\mathrm{S}}_7=2059\\ \mathrm{Thus},\text{sum of 7 terms of G.P. is 2059.}\end{array}$

Q.39 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Ans

$\begin{array}{l}\mathrm{Let}\text{G.P. be:}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}{\text{, ar}}^{\text{5}}\text{,}...\\ \text{Then according to first condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} a}+\mathrm{ar}=-4\\ \mathrm{or}\mathrm{a}(1+\mathrm{r})=-4\\ \mathrm{or}\mathrm{a}=\frac{-4}{1+\mathrm{r}}...\left(\mathrm{i}\right)\\ \text{Then according to second condition,}\\ {\text{T}}_{\text{5}}={\text{4T}}_{\text{3}}\\ \Rightarrow {\text{\hspace{0.17em}ar}}^{\text{4}}=4\left({\text{ar}}^{\text{2}}\right)\\ \Rightarrow {\mathrm{r}}^2=4\\ \Rightarrow \mathrm{r}=\pm 2\\ \mathrm{Putting}\text{r}=2\text{in equation}\left(\mathrm{i}\right),\text{we get}\end{array}$ $\begin{array}{l}\mathrm{a}=\frac{-4}{1+2}\\ =-\frac{4}{3}\\ \mathrm{Putting}\text{r}=2\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{a}=\frac{-4}{1-2}\\ =\frac{-4}{-1}\\ =4\\ \mathrm{Therefore},\text{G.P. when a}=-\frac{4}{3}\text{and r}=\text{2}\\ -\frac{4}{3}\text{,}-\frac{4}{3}\left(2\right)\text{,}-\frac{4}{3}{\left(2\right)}^{\text{2}}\text{,}-\frac{4}{3}{\left(2\right)}^{\text{3}}\text{,}-\frac{4}{3}{\left(2\right)}^{\text{4}}\text{,}-\frac{4}{3}{\left(2\right)}^{\text{5}}\text{,}...\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{4}{3},-\frac{8}{3},-\frac{16}{3},-\frac{32}{3},-\frac{64}{3},-\frac{128}{3},...\\ \mathrm{Therefore},\text{G.P. when a}=4\text{and r}=-\text{2}\\ 4\text{,}4(-2)\text{,}4{(-2)}^{\text{2}}\text{,}4{(-2)}^{\text{3}}\text{,}4{(-2)}^{\text{4}}\text{,}4{(-2)}^{\text{5}}\text{,}...\\ \text{or\hspace{0.17em}\hspace{0.17em}}4,-8,16,-32,...\\ \mathrm{Therefore},\text{the required G.P. is}\\ -\frac{4}{3},-\frac{8}{3},-\frac{16}{3},-\frac{32}{3},-\frac{64}{3},-\frac{128}{3},...\\ \mathrm{or}4,-8,16,-32,...\end{array}$

Q.40 If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Ans

$\mathbf{Let}\mathbf{\text{first term of G.P. be ‘a’ and common ratio be ‘r’.}}$ $\begin{array}{l}{\text{\hspace{0.17em} T}}_{\text{4}}=\text{x}\\ \Rightarrow {\mathrm{ar}}^3=\mathrm{x}\\ {\text{T}}_{\text{10}}=\text{y}\\ \Rightarrow {\mathrm{ar}}^9=\mathrm{y}\\ {\text{T}}_{\text{16}}=\text{z}\\ \Rightarrow {\mathrm{ar}}^{15}=\mathrm{z}\\ \mathrm{Now},\frac{\mathrm{y}}{\mathrm{x}}=\frac{{\mathrm{ar}}^9}{{\mathrm{ar}}^3}\\ ={\mathrm{r}}^6\\ \mathrm{and}\frac{\mathrm{z}}{\mathrm{y}}=\frac{{\mathrm{ar}}^{15}}{{\mathrm{ar}}^9}\\ ={\mathrm{r}}^6\\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{y}}\\ \Rightarrow \mathrm{x},\mathrm{y},\mathrm{z}\text{are in G.P.}\end{array}$

Q.41 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{S}}_{\mathrm{n}}=\text{8}+\text{88}+\text{888}+\text{8888}+\dots \\ =8(1+11+111+1111+...)\\ =\frac{8}{9}(9+99+999+9999+...)\\ =\frac{8}{9}\{(10-1)+({10}^2-1)+({10}^3-1)+({10}^4-1)+...\}\\ =\frac{8}{9}\{10+{10}^2+{10}^3+{10}^4+...-1-1-1-1-...\mathrm{n}\text{times}\}\end{array}$ $\begin{array}{l}=\frac{8}{9}\{\frac{10({10}^{\mathrm{n}}-1)}{10-1}-\mathrm{n}\}[âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}]\\ =\frac{8}{9}\{\frac{10({10}^{\mathrm{n}}-1)}{9}-\mathrm{n}\}\\ =\frac{80}{81}({10}^{\mathrm{n}}-1)-\frac{8}{9}\mathrm{n}\end{array}$

Q.42

$\begin{array}{l}\mathbf{\text{Find the sum of the products of the corresponding terms}}\\ \mathbf{\text{of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2,}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{Sum of the products of the corresponding terms of the sequences}\\ \text{2, 4, 8, 16, 32 and 128, 32, 8, 2,}\frac{1}{2}\\ =2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}\\ =256+128+64+32+16\\ =\frac{256\left\{1-{\left(\frac{1}{2}\right)}^5\right\}}{1-\frac{1}{2}}=\frac{256\left(\frac{32-1}{32}\right)}{\frac{1}{2}}\\ =256\times 2\times \frac{31}{32}=496\end{array}$

Q.43 Show that the products of the corresponding terms of the sequences a, ar, ar², … ar^{n – 1} and A, AR, AR², … AR^n–1 form a G.P, and find the common ratio.

Ans

$\begin{array}{l}\text{The products of the corresponding terms of the sequences}\\ \text{a},\text{ar},{\text{ar}}^{\text{2}},\dots {{\text{ar}}^{\text{n}}}^{–\text{1}}\mathrm{}\text{and A},\text{AR},{\text{AR}}^{\text{2}},\dots {\text{AR}}^{\text{n}–\text{1}}\\ =\text{aA},\text{aARr},{\text{aAR}}^2{\text{r}}^{\text{2}},\dots {{\text{,aAR}}^{\text{n}}}^{–\text{1}}{{\text{r}}^{\text{n}}}^{–\text{1}}\\ \mathrm{Common}\text{ratio}=\frac{\text{aARr}}{\text{aA}}=\text{Rr}\\ \mathrm{Common}\text{ratio}=\frac{{\text{aAR}}^2{\text{r}}^{\text{2}}}{\text{aARr}}=\text{Rr}\\ \mathrm{Since},\text{common ratio is same i.e., rR.}\\ \text{So, aA},\text{aARr},{\text{aAR}}^2{\text{r}}^{\text{2}},\dots {{\text{,aAR}}^{\text{n}}}^{–\text{1}}{{\text{r}}^{\text{n}}}^{–\text{1}}\text{is in G.P.}\end{array}$

Q.44 Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.

Ans

$\begin{array}{l}{\text{Let four numbers of G.P. are a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{.}\\ \text{According to first condition,}\\ {\text{\hspace{0.17em}\hspace{0.17em} T}}_{\text{3}}-{\text{T}}_{\text{1}}=9\\ \Rightarrow {\mathrm{ar}}^2-\mathrm{a}=9\\ \Rightarrow \mathrm{a}\left({\mathrm{r}}^2-1\right)=9...\left(\mathrm{i}\right)\\ \text{According to second condition,}\\ {\text{T}}_{\text{2}}-{\text{T}}_{\text{4}}=18\\ \Rightarrow \mathrm{ar}-{\mathrm{ar}}^3=18\\ \Rightarrow \mathrm{ar}\left(1-{\mathrm{r}}^2\right)=18\\ \Rightarrow \mathrm{ar}\left({\mathrm{r}}^2-1\right)=-18...\left(\mathrm{ii}\right)\\ \text{Dividing equation}\left(\text{ii}\right)\text{by equation\hspace{0.17em}}\left(\text{i}\right)\text{, we get}\\ \frac{\mathrm{ar}\left({\mathrm{r}}^2-1\right)}{\mathrm{a}\left({\mathrm{r}}^2-1\right)}=-\frac{18}{9}\\ \Rightarrow \mathrm{r}=-2\\ \text{Putting r=}-\text{2, in equation}\left(\text{i}\right)\text{, we get}\\ \mathrm{a}\left\{{\left(-2\right)}^2-1\right\}=9\\ \mathrm{a}=\frac{9}{3}\\ =3\\ \text{Thus, the four terms of G.P. are:}\\ \text{3, 3}\left(-2\right)\text{, 3}{\left(-2\right)}^{\text{2}}\text{, 3}{\left(-2\right)}^{\text{3}}\\ \mathrm{i}.\mathrm{e}.,3,-6,12,-24\end{array}$

Q.45

$\begin{array}{l}{\mathbf{\text{If the p}}}^{\mathbf{\text{th}}}{\mathbf{\text{, q}}}^{\mathbf{\text{th}}}{\mathbf{\text{\hspace{0.17em}and r}}}^{\mathbf{\text{th}}}\mathbf{\text{terms of a G.P. are a and b, respectively.}}\\ {\mathbf{\text{Prove that\hspace{0.17em} a}}}^{\mathbf{\text{q}}\mathbf{-}\mathbf{\text{r}}}{\mathbf{\text{\hspace{0.17em}b}}}^{\mathbf{\text{r}}\mathbf{-}\mathbf{\text{p}}}{\mathbf{\text{\hspace{0.17em}c}}}^{\mathbf{\text{p}}\mathbf{-}\mathbf{\text{q}}}\mathbf{\text{=1.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of G.P. is A and common ratio is R.}\\ \mathrm{Given}\text{that:}\\ {\text{T}}_{\text{p}}=\text{a}\Rightarrow {\mathrm{AR}}^{\mathrm{p}-1}=\mathrm{a}\\ {\text{T}}_{\text{q}}=\text{b}\Rightarrow {\mathrm{AR}}^{\mathrm{q}-1}=\mathrm{b}\\ {\text{T}}_{\text{r}}=\text{c}\Rightarrow {\mathrm{AR}}^{\mathrm{r}-1}=\mathrm{c}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{a}}^{\mathrm{q}–\mathrm{r}}{\mathrm{b}}^{\mathrm{r}–\mathrm{p}}{\mathrm{c}}^{\mathrm{p}–\mathrm{q}}\\ ={\left({\mathrm{AR}}^{\mathrm{p}-1}\right)}^{\mathrm{q}–\mathrm{r}}{\left({\mathrm{AR}}^{\mathrm{q}-1}\right)}^{\mathrm{r}–\mathrm{p}}{\left({\mathrm{AR}}^{\mathrm{r}-1}\right)}^{\mathrm{p}–\mathrm{q}}\\ ={\mathrm{A}}^{\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q}}{\mathrm{R}}^{(\mathrm{p}-1)(\mathrm{q}-\mathrm{r})+(\mathrm{q}-1)(\mathrm{r}-\mathrm{p})(\mathrm{r}-1)(\mathrm{p}-\mathrm{q})}\\ ={\mathrm{A}}^0{\mathrm{R}}^{\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{qp}-\mathrm{r}+\mathrm{p}+\mathrm{rp}-\mathrm{rq}-\mathrm{p}+\mathrm{q}}\\ ={\mathrm{A}}^0{\mathrm{R}}^0\\ =1\times 1\\ =1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.46 If the first and the n^th term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\text{a}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}}^{\text{th}}\text{term of G.P.}=\text{b}\\ \text{Let common ratio}=\mathrm{r}\\ \mathrm{The}\text{G.P. is}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}\text{,}...\text{, b}\\ \text{and b}={\text{ar}}^{\text{n-1}}\\ \text{Product of n terms}=\text{a}\times \text{ar}\times {\text{ar}}^{\text{2}}\times {\text{ar}}^{\text{3}}\times {\text{ar}}^{\text{4}}\times ...\times \text{b}\\ =\text{a}\times \text{ar}\times {\text{ar}}^{\text{2}}\times {\text{ar}}^{\text{3}}\times {\text{ar}}^{\text{4}}\times ...\times {\text{ar}}^{\text{n-1}}\end{array}$ $\begin{array}{l}={\mathrm{a}}^{\mathrm{n}}\times {\mathrm{r}}^{1+2+3+4+...+(\mathrm{n}-1)}\\ \mathrm{P}={\mathrm{a}}^{\mathrm{n}}\times {\mathrm{r}}^{\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}}\\ \mathrm{P}={\mathrm{a}}^{\mathrm{n}}\times {\mathrm{r}}^{\frac{\mathrm{n}(\mathrm{n}-1)}{2}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {\text{P}}^{\text{2}}={\mathrm{a}}^{2\mathrm{n}}{\mathrm{r}}^{\mathrm{n}(\mathrm{n}-1)}\\ ={\left({\mathrm{a}}^2{\mathrm{r}}^{\mathrm{n}-1}\right)}^{\mathrm{n}}\\ ={(\mathrm{a}\times {\mathrm{ar}}^{\mathrm{n}-1})}^{\mathrm{n}}\\ {\text{\hspace{0.17em}P}}^{\text{2}}={\left(\mathrm{ab}\right)}^{\mathrm{n}}[âˆ\mu \mathrm{b}={\mathrm{ar}}^{\mathrm{n}-1}]\\ \mathrm{Hence}\text{Proved.}\end{array}$

Q.47

$\begin{array}{l}\mathbf{\text{Show that the ratio of the sum of first n terms of a G.P.}}\\ \mathbf{\text{to the sum of terms from \hspace{0.17em}}}{\left(\text{n+1}\right)}^{\mathbf{\text{th}}}\mathbf{\text{to}}{\left(\text{2n}\right)}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{{\mathbf{\text{r}}}^{\mathbf{\text{n}}}}\mathbf{.}\end{array}$

Ans

$\begin{array}{l}\text{LetfirsttermofG.P.isaandcommonratioisr.Then,}\\ \text{Sumoffirstnterms}\left({\text{S}}_{\text{n}}\right)\text{=}\frac{\text{a}\left({\text{r}}^{\text{n}}\text{-1}\right)}{\left(\text{r-1}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sumof2nterms}\left({\text{S}}_{\text{2n}}\right)\text{=}\frac{\text{a}\left({\text{r}}^{\text{2n}}\text{-1}\right)}{\left(\text{r-1}\right)}\\ \text{Sumoftermsfrom\hspace{0.17em}}{\left(\text{n+1}\right)}^{\text{th}}\text{to}{\left(\text{2n}\right)}^{\text{th}}\text{term}\\ {\text{=S}}_{\text{2n}}{\text{-S}}_{\text{n}}\end{array}$ $\begin{array}{l}\text{=}\frac{\text{a}\left({\text{r}}^{\text{2n}}\text{-1}\right)}{\left(\text{r-1}\right)}\text{–}\frac{\text{a}\left({\text{r}}^{\text{n}}\text{-1}\right)}{\left(\text{r-1}\right)}\\ \text{=}\frac{\text{a}}{\left(\text{r-1}\right)}\left({\text{r}}^{\text{2n}}{\text{-1-r}}^{\text{n}}\text{+1}\right)\\ \text{=}\frac{{\text{ar}}^{\text{n}}}{\left(\text{r-1}\right)}\left({\text{r}}^{\text{n}}\text{-1}\right)\\ \text{The required ratio=}\frac{{\text{S}}_{\text{n}}}{{\text{S}}_{\text{2n}}{\text{-S}}_{\text{n}}}\\ \text{=}\frac{\frac{\text{a}\left({\text{r}}^{\text{n}}\text{-1}\right)}{\left(\text{r-1}\right)}}{\frac{{\text{ar}}^{\text{n}}}{\left(\text{r-1}\right)}\left({\text{r}}^{\text{n}}\text{-1}\right)}\\ \text{=}\frac{\text{1}}{{\text{r}}^{\text{n}}}\\ \text{Thus,\hspace{0.17em}\hspace{0.17em}the ratio of the sum of first n terms of a G.P. to}\\ \text{the sum of terms from \hspace{0.17em}}{\left(\text{n+1}\right)}^{\text{th}}\text{to}{\left(\text{2n}\right)}^{\text{th}}\text{term is}\frac{\text{1}}{{\text{r}}^{\text{n}}\text{.}}\end{array}$

Q.48 If a, b, c and d are in G.P. show that (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)².

Ans

$\begin{array}{l}\mathrm{Since},\text{a, b, c, d are in G.P.}\\ \text{So,}\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ \Rightarrow \mathrm{b}=\mathrm{ar},\text{c}=\mathrm{br}={\mathrm{ar}}^2\\ \mathrm{and}\text{d}=\mathrm{cr}={\mathrm{ar}}^4\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=({\text{a}}^{\text{2}}+{\text{b}}^{\text{2}}+{\text{c}}^{\text{2}})({\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}+{\text{d}}^{\text{2}})\end{array}$ $\begin{array}{l}=({\text{a}}^{\text{2}}+{\text{a}}^{\text{2}}{\mathrm{r}}^2+{\text{a}}^{\text{2}}{\mathrm{r}}^4)({\text{a}}^{\text{2}}{\mathrm{r}}^2+{\text{a}}^{\text{2}}{\mathrm{r}}^4+{\text{a}}^{\text{2}}{\mathrm{r}}^6)\\ ={\mathrm{a}}^2(1+{\mathrm{r}}^2+{\mathrm{r}}^4)\times {\text{a}}^{\text{2}}{\mathrm{r}}^2(1+{\mathrm{r}}^2+{\mathrm{r}}^4)\\ ={\mathrm{a}}^4{\mathrm{r}}^2{(1+{\mathrm{r}}^2+{\mathrm{r}}^4)}^2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.={(\text{ab}+\text{bc}+\text{cd})}^{\text{2}}\\ ={(\text{a}\times \mathrm{ar}+\mathrm{ar}\times {\mathrm{ar}}^2+{\mathrm{ar}}^2\times {\mathrm{ar}}^4)}^{\text{2}}\\ ={\mathrm{a}}^4{\mathrm{r}}^2{(1+{\mathrm{r}}^2+{\mathrm{r}}^4)}^2\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Therefore},\\ ({\text{a}}^{\text{2}}+{\text{b}}^{\text{2}}+{\text{c}}^{\text{2}})({\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}+{\text{d}}^{\text{2}})={(\text{ab}+\text{bc}+\text{cd})}^{\text{2}}\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.49 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Ans

$\begin{array}{l}\mathrm{Let}\text{two numbers between 3 and 81 are x and y in such a way}\\ \text{that 3, x, y, 81 are in G.P.}\\ \text{So, first term of G.P.}=\text{3}\\ \mathrm{Common}\text{difference}=\mathrm{r}\\ 4^{\mathrm{th}}\mathrm{term}\text{of G.P.}={\mathrm{ar}}^3\\ \Rightarrow 3{\mathrm{r}}^3=81\\ \Rightarrow {\mathrm{r}}^3=\frac{81}{3}\\ \Rightarrow \mathrm{r}=\sqrt[3]{27}\\ =3\end{array}$ $\begin{array}{l}\therefore \mathrm{x}={\mathrm{T}}_2\\ =\mathrm{ar}\\ =3\times 3\\ =9\\ \mathrm{andy}={\mathrm{T}}_3\\ ={\mathrm{ar}}^2\\ =3\times 3^2\\ =27\\ \mathrm{Thus},\text{two numbers between 3 and 81 are 9 and 27.}\end{array}$

Q.50

$\mathbf{\text{Find the value of n so that\hspace{0.17em}}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n+1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n+1}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}\mathbf{\text{may be the geometric mean between a and b.}}$

Ans

$\begin{array}{l}\mathrm{Geometric}\text{mean between a and b}=\sqrt{\mathrm{ab}}\\ \frac{{\text{a}}^{\text{n}+1}+{\text{b}}^{\text{n}+1}}{{\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}}=\sqrt{\mathrm{ab}}\\ \Rightarrow {\left(\mathrm{ab}\right)}^{\frac{1}{2}}\left({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}\right)={\text{a}}^{\text{n}+1}+{\text{b}}^{\text{n}+1}\\ \Rightarrow {\text{a}}^{\text{n}+\frac{1}{2}}{\text{b}}^{\frac{1}{2}}+{\text{a}}^{\frac{1}{2}}{\text{b}}^{\text{n}+\frac{1}{2}}={\text{a}}^{\text{n}+1}+{\text{b}}^{\text{n}+1}\\ \Rightarrow {\text{a}}^{\text{n}+\frac{1}{2}}{\text{b}}^{\frac{1}{2}}-{\text{a}}^{\text{n}+1}={\text{b}}^{\text{n}+1}-{\text{a}}^{\frac{1}{2}}{\text{b}}^{\text{n}+\frac{1}{2}}\\ \Rightarrow {\text{a}}^{\text{n}+\frac{1}{2}}\left({\text{b}}^{\frac{1}{2}}-{\text{a}}^{\frac{1}{2}}\right)={\text{b}}^{\text{n}+\frac{1}{2}}\left({\text{b}}^{\frac{1}{2}}-{\text{a}}^{\frac{1}{2}}\right)\\ \Rightarrow {\text{a}}^{\text{n}+\frac{1}{2}}={\text{b}}^{\text{n}+\frac{1}{2}}\\ \Rightarrow {\left(\frac{\text{a}}{\text{b}}\right)}^{\text{n}+\frac{1}{2}}=1\\ \Rightarrow {\left(\frac{\text{a}}{\text{b}}\right)}^{\text{n}+\frac{1}{2}}={\left(\frac{\text{b}}{\text{a}}\right)}^0\\ \Rightarrow \text{n}+\frac{1}{2}=0\\ \Rightarrow \text{n}=-\frac{1}{2}\end{array}$

Q.51

$\begin{array}{l}\mathbf{\text{The sum of two numbers is 6 times their geometric mean, show that}}\\ \mathbf{\text{numbers are in the ratio}}\left(\text{3+2}\sqrt{\text{2}}\right)\mathbf{\text{:}}(\text{3}-\text{2}\sqrt{\text{2}})\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{two numbers be a and b.}\\ \text{G.M. of a and b}=\sqrt{\mathrm{ab}}\\ \mathrm{Then},\text{according to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}a}+\text{b}=6\sqrt{\mathrm{ab}}...\left(\mathrm{i}\right)\\ âˆ\mu {(\text{a}-\text{b})}^2={(\text{a}+\text{b})}^2-4\mathrm{ab}\\ \Rightarrow {(\text{a}-\text{b})}^2={\left(6\sqrt{\mathrm{ab}}\right)}^2-4\mathrm{ab}\\ =36\mathrm{ab}-4\mathrm{ab}\\ =32\mathrm{ab}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} a}-\text{b}=\sqrt{32\mathrm{ab}}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} a}-\text{b}=4\sqrt{2}\sqrt{\mathrm{ab}}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}\mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{a}=(3+2\sqrt{2})\sqrt{\mathrm{ab}}\text{and b}=(3-2\sqrt{2})\sqrt{\mathrm{ab}}\\ \mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ratio of the numbers}=\frac{(3+2\sqrt{2})\sqrt{\mathrm{ab}}}{(3-2\sqrt{2})\sqrt{\mathrm{ab}}}\\ =(3+2\sqrt{2}):(3-2\sqrt{2})\end{array}$

Q.52

$\begin{array}{l}\mathbf{\text{If A and G be A.M. and G.M., respectively between two positive numbers,}}\\ \mathbf{\text{prove that the numbers are\hspace{0.17em}\hspace{0.17em}A±}}\sqrt{\left(\text{A+G}\right)(\text{A}-\text{G})}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{Let two numbers are a and b.}\\ \text{Then, A.M.}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \Rightarrow \mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \Rightarrow \mathrm{a}+\mathrm{b}=2\mathrm{A}...\left(\mathrm{i}\right)\\ âˆ\mu \mathrm{G}.\mathrm{M}.=\sqrt{\mathrm{ab}}\\ \Rightarrow \mathrm{G}=\sqrt{\mathrm{ab}}\\ \Rightarrow {\mathrm{G}}^2=\mathrm{ab}...\left(\mathrm{ii}\right)\\ âˆ\mu \mathrm{a}-\mathrm{b}=\sqrt{{\left(\mathrm{a}+\mathrm{b}\right)}^2-4\mathrm{ab}}\\ =\sqrt{{\left(2\mathrm{A}\right)}^2-4{\mathrm{G}}^2}\\ \mathrm{a}-\mathrm{b}=2\sqrt{{\mathrm{A}}^2-{\mathrm{G}}^2}...\left(\mathrm{iii}\right)\\ \text{Solving equation}\left(\text{i}\right)\text{and equation}\left(\text{iii}\right)\text{, we get}\\ \text{a}=\mathrm{A}+\sqrt{{\mathrm{A}}^2-{\mathrm{G}}^2}\text{and b}=\mathrm{A}-\sqrt{{\mathrm{A}}^2-{\mathrm{G}}^2}\\ \text{Therefore, the two numbers are:}\\ \mathrm{A}\pm \sqrt{\left(\mathrm{A}+\mathrm{G}\right)\left(\mathrm{A}-\mathrm{G}\right)}.\end{array}$

Q.53 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?

Ans

$\begin{array}{l}\mathrm{Originally}\text{present bacteria in a culture}=30\\ \mathrm{Number}\text{of bacteria after one hour}=60\\ \mathrm{Since},\text{number of bacteria doubles every hour.}\\ \text{So, common ratio}\left(\mathrm{r}\right)\text{in G.P.}=2\\ \mathrm{First}\text{term}\left(\mathrm{a}\right)\text{in G.P.}=30\\ \mathrm{Number}{\text{of bacteria after 2}}^{\text{nd}}\text{hour}\\ ={\text{ar}}^{\text{2}}\\ =30\times 2^2\\ =120\\ \mathrm{Number}{\text{of bacteria after 4}}^{\text{th}}\text{hour}\\ ={\text{ar}}^{\text{4}}\\ =30\times 2^4\\ =480\end{array}$ $\begin{array}{l}\mathrm{Number}{\text{of bacteria after n}}^{\text{th}}\text{hour}\\ ={\text{ar}}^{\text{n}}\\ =30\times 2^{\mathrm{n}}\\ =30\left(2^{\mathrm{n}}\right)\end{array}$

Q.54 What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Ans

$\begin{array}{l}\text{Principal amount deposited in the bank}=₹500\\ \mathrm{Rate}\text{of interest in the bank}=10\mathrm{\%}\mathrm{p}.\mathrm{a}.\\ \mathrm{Number}\text{of years for amount deposited}=10\text{years}\\ \mathrm{Amount}\text{after one year in the bank}=500(1+\frac{10}{100})\\ =500\left(\frac{11}{10}\right)\\ \mathrm{Amount}\text{after two years in the bank}=500{(1+\frac{10}{100})}^2\\ =500{\left(\frac{11}{10}\right)}^2\\ \mathrm{Amount}\text{after three years in the bank}=500{(1+\frac{10}{100})}^3\\ =500{\left(\frac{11}{10}\right)}^3\\ \mathrm{So},\text{G.P. formed by using amount obtained from bank is}\end{array}$ $\begin{array}{l}500\left(\frac{11}{10}\right),500{\left(\frac{11}{10}\right)}^2,500{\left(\frac{11}{10}\right)}^3,...\\ \mathrm{First}\text{term of G.P.}=500\left(\frac{11}{10}\right)\\ \mathrm{Common}\text{ratio of G.P.}=\frac{11}{10}\\ \mathrm{Number}\text{of years}=10\\ \mathrm{Since},{\text{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ \Rightarrow {\text{T}}_{10}=500\left(\frac{11}{10}\right)\times {\left(\frac{11}{10}\right)}^{10-1}\\ =500{(1.1)}^{10}\\ \mathrm{Thus},₹\text{\hspace{0.17em}500 amounts to}₹500{(1.1)}^{10}\text{in 10 years after its}\\ \text{deposit in a bank.}\end{array}$

Q.55 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Ans

$\begin{array}{l}\mathrm{Let}\text{roots of a quadratic equation be a and b.}\\ \text{A.M. of a and b}=8\\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}=8\\ \Rightarrow \mathrm{a}+\mathrm{b}=16...\left(\mathrm{i}\right)\\ \mathrm{G}.\mathrm{M}.\text{of a and b}=5\\ \Rightarrow \sqrt{\mathrm{ab}}=5\\ \Rightarrow \mathrm{ab}=25...\left(\mathrm{ii}\right)\\ \mathrm{Quadratic}\text{equation, whose roots are a and b,}\end{array}$ $\begin{array}{l}(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})=0\\ {\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}=0\\ {\mathrm{x}}^2-16\mathrm{x}+25=0\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right)\right]\\ \mathrm{Thus},\text{the required quadratic equation is \hspace{0.17em}}{\mathrm{x}}^2-16\mathrm{x}+25=0.\end{array}$

Q.56 Find the sum to n terms of each of the series in Exercises 1 to 7.

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +…
2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
3. 3 × 1² + 5 × 2² + 7 × 3² + …
4.

\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\mathrm{\dots }

5. 5² + 6² + 7² + … + 20<sup>2
6. 3 × 8 + 6 × 11 + 9 × 14 + …
7. 12 + (12 + 22) + (12 + 22 + 32) + …</sup>

Ans

$\begin{array}{l}1.\mathrm{Let}{\text{\hspace{0.17em} S}}_{\text{n}}=\text{1}\times \text{2}+\text{2}\times \text{3}+\text{3}\times \text{4}+\text{4}\times \text{5}+...+\mathrm{n}\times \left(\mathrm{n}+1\right)\\ {\mathrm{a}}_{\mathrm{n}}=\mathrm{n}\times \left(\mathrm{n}+1\right)\\ {\mathrm{S}}_{\mathrm{n}}=\sum \mathrm{n}\times \left(\mathrm{n}+1\right)\\ =\sum {\mathrm{n}}^2+\sum \mathrm{n}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{2\mathrm{n}+1}{3}+1\right)\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{2\mathrm{n}+4}{3}\right)\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\times 2\left(\frac{\mathrm{n}+2}{3}\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{1}{3}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\\ \text{2.}\mathrm{Let}{\text{\hspace{0.17em} S}}_{\text{n}}=\text{1}\times \text{2}\times \text{3}+\text{2}\times \text{3}\times \text{4}+\text{3}\times \text{4}\times \text{5}+...+\mathrm{n}\times \left(\mathrm{n}+1\right)\times \left(\mathrm{n}+2\right)\\ {\mathrm{a}}_{\mathrm{n}}=\mathrm{n}\times \left(\mathrm{n}+1\right)\times \left(\mathrm{n}+2\right)\\ {\mathrm{S}}_{\mathrm{n}}=\sum \mathrm{n}\times \left(\mathrm{n}+1\right)\times \left(\mathrm{n}+2\right)\\ =\sum \left({\mathrm{n}}^3+3{\mathrm{n}}^2+2\mathrm{n}\right)\\ =\sum {\mathrm{n}}^3+3\sum {\mathrm{n}}^2+2\sum \mathrm{n}\\ ={\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^2+3.\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+2.\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ =\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+\frac{2\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\left(2\mathrm{n}+1\right)+2\right\}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{{\mathrm{n}}^2+\mathrm{n}+4\mathrm{n}+2+4}{2}\right\}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{{\mathrm{n}}^2+5\mathrm{n}+6}{2}\right)\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)}{4}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{1}{4}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)\\ 3.{\mathrm{a}}_{\mathrm{n}}=\left(2\mathrm{n}+1\right)\times {\mathrm{n}}^2\\ {\mathrm{S}}_{\mathrm{n}}=\sum \left(2{\mathrm{n}}^3+{\mathrm{n}}^2\right)\\ =2\sum {\mathrm{n}}^3+\sum {\mathrm{n}}^2\\ =2{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^2+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ =2.\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ =\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\mathrm{n}\left(\mathrm{n}+1\right)+\frac{\left(2\mathrm{n}+1\right)}{3}\right\}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3\mathrm{n}\left(\mathrm{n}+1\right)+\left(2\mathrm{n}+1\right)}{3}\right\}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3{\mathrm{n}}^2+3\mathrm{n}+2\mathrm{n}+1}{3}\right\}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3{\mathrm{n}}^2+5\mathrm{n}+1}{3}\right\}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{1}{6}\mathrm{n}\left(\mathrm{n}+1\right)\left(3{\mathrm{n}}^2+5\mathrm{n}+1\right)\\ 4.\mathrm{Let}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{\mathrm{n}\left(\mathrm{n}+1\right)}\\ =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right)\\ =\frac{1}{1}-\frac{1}{\mathrm{n}+1}\\ =\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\\ =\frac{\mathrm{n}}{\mathrm{n}+1}\\ \text{5.}{5}^2+6^2+7^2+...+{20}^2\\ =1^2+2^2+3^2+4^2+5^2+6^2+7^2+...+{20}^2-\left(1^2+2^2+3^2+4^2\right)\\ =\sum {20}^2-\sum 4^2\\ =\frac{20\left(20+1\right)\left(2\times 20+1\right)}{6}-\frac{4\left(4+1\right)\left(2\times 4+1\right)}{6}\\ =\frac{20\times 21\times 41}{6}-\frac{4\times 5\times 9}{6}\\ =\frac{17220-180}{6}\\ =\frac{17040}{6}\\ =2840\\ 6.{\text{Let\hspace{0.17em}S}}_{\text{n}}=\text{3}\times \text{8}+\text{6}\times \text{11}+\text{9}\times \text{14}+...\\ {\mathrm{T}}_{\mathrm{n}}=\left(3,6,9,...{\mathrm{n}}^{\mathrm{th}}\mathrm{term}\right)\left(8,11,14,...{\text{n}}^{\text{th}}\text{term}\right)\\ =\left\{3+\left(\mathrm{n}-1\right)3\right\}\left\{8+\left(\mathrm{n}-1\right)3\right\}\\ =\left(3\mathrm{n}\right)\left(3\mathrm{n}+5\right)\\ =9{\mathrm{n}}^2+15\mathrm{n}\\ {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ =\sum \left(9{\mathrm{n}}^2+15\mathrm{n}\right)\\ =9\sum {\mathrm{n}}^2+15\sum \mathrm{n}\\ =9\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+15\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ =3\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+15\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ =\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\left(2\mathrm{n}+1\right)+5\right\}\\ =\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(2\mathrm{n}+6\right)\\ =3\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+3\right)\\ 7.\mathrm{Let}{\mathrm{S}}_{\mathrm{n}}={\text{1}}^{\text{2}}+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}\right)+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}\right)+...+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+...+{\mathrm{n}}^2\right)\\ \mathrm{Here},{\text{\hspace{0.17em}T}}_{\text{n}}={\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+...+{\mathrm{n}}^2\\ =\sum {\mathrm{n}}^2\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ =\frac{1}{6}\mathrm{n}\left(2{\mathrm{n}}^2+3\mathrm{n}+1\right)\\ =\frac{1}{6}\left(2{\mathrm{n}}^3+3{\mathrm{n}}^2+\mathrm{n}\right)\\ {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ =\frac{1}{6}\sum \left(2{\mathrm{n}}^3+3{\mathrm{n}}^2+\mathrm{n}\right)\\ =\frac{1}{6}\left(2\sum {\mathrm{n}}^3+3\sum {\mathrm{n}}^2+\sum \mathrm{n}\right)\\ =\frac{1}{6}\left[2{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^2+3\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]\\ =\frac{1}{6}\left\{\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}\\ =\frac{1}{6}\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\mathrm{n}\left(\mathrm{n}+1\right)+\left(2\mathrm{n}+1\right)+1\right\}\\ =\frac{1}{6}\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left({\mathrm{n}}^2+\mathrm{n}+2\mathrm{n}+2\right)\\ =\frac{1}{12}\mathrm{n}\left(\mathrm{n}+1\right)\left({\mathrm{n}}^2+3\mathrm{n}+2\right)\\ =\frac{1}{12}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}{\left(\mathrm{n}+1\right)}^2\left(\mathrm{n}+2\right)}{12}\end{array}$

Q.57 Find the sum to n terms of the series in whose n^th terms is given by:

1. n (n + 1) (n + 4).
2. n² + 2<sup>n
3. (2n – 1)2</sup>

Ans

1.

${\text{Let\hspace{0.17em}\hspace{0.17em}T}}_{\text{n}}=\text{n}(\text{n}+\text{1})(\text{n}+\text{4})$ $\begin{array}{l}=\mathrm{n}({\mathrm{n}}^2+5\mathrm{n}+4)\\ ={\mathrm{n}}^3+5{\mathrm{n}}^2+4\mathrm{n}\\ {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ =\sum ({\mathrm{n}}^3+5{\mathrm{n}}^2+4\mathrm{n})\\ =\sum {\mathrm{n}}^3+5\sum {\mathrm{n}}^2+4\sum \mathrm{n}\\ ={\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}}^2+\frac{5\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\frac{4\mathrm{n}(\mathrm{n}+1)}{2}\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}+5.\frac{(2\mathrm{n}+1)}{3}+4\}\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left\{\frac{3\mathrm{n}(\mathrm{n}+1)+10(2\mathrm{n}+1)+24}{6}\right\}\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{3{\mathrm{n}}^2+3\mathrm{n}+20\mathrm{n}+10+24}{6}\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{12}(3{\mathrm{n}}^2+23\mathrm{n}+34)\end{array}$

2.

$\begin{array}{l}\mathrm{Let}{\text{\hspace{0.17em}\hspace{0.17em}T}}_{\text{n}}={\text{n}}^{\text{2}}+{\text{2}}^{\text{n}}\\ {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ =\sum ({\text{n}}^{\text{2}}+{\text{2}}^{\text{n}})\\ =\sum {\text{n}}^{\text{2}}+\sum {\text{2}}^{\text{n}}\\ =\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+(2+2^2+2^3+2^4+...+2^{\mathrm{n}})\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\frac{2(2^{\mathrm{n}}-1)}{2-1}[âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}]\\ =\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+2(2^{\mathrm{n}}-1)\\ \therefore {\mathrm{S}}_{\mathrm{n}}=\frac{1}{6}\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)+2(2^{\mathrm{n}}-1)\end{array}$

3.

$\begin{array}{l}\mathrm{Let}{\mathrm{t}}_{\mathrm{n}}={(\text{2n}-\text{1})}^{\text{2}}\\ =4{\mathrm{n}}^2-4\mathrm{n}+1\\ {\mathrm{S}}_{\mathrm{n}}=\sum (4{\mathrm{n}}^2-4\mathrm{n}+1)\\ =4\sum {\mathrm{n}}^2-4\sum \mathrm{n}+\sum 1\\ =4\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}-4\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\\ =2\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{3}-2\mathrm{n}(\mathrm{n}+1)+\mathrm{n}\\ =\frac{2\mathrm{n}(2{\mathrm{n}}^2+3\mathrm{n}+1)}{3}-2{\mathrm{n}}^2-2\mathrm{n}+\mathrm{n}\\ =\frac{2(2{\mathrm{n}}^3+3{\mathrm{n}}^2+\mathrm{n})}{3}-2{\mathrm{n}}^2-\mathrm{n}\\ =\frac{4{\mathrm{n}}^3+6{\mathrm{n}}^2+2\mathrm{n}-6{\mathrm{n}}^2-3\mathrm{n}}{3}\\ =\frac{4{\mathrm{n}}^3-\mathrm{n}}{3}\end{array}$ $\begin{array}{l}=\frac{\mathrm{n}}{3}(4{\mathrm{n}}^2-1)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{3}(2\mathrm{n}-1)(2\mathrm{n}+1)\end{array}$

Q.58

$\begin{array}{l}\text{Find the sum to infinity in each of the following Geometric Progression.}\\ \text{1. 1,}\frac{\text{1}}{\text{3}}\text{,}\frac{\text{1}}{\text{9}}\text{,}...\\ \text{2. 6, 1.2, 0.24,}...\\ \text{3. 5,}\frac{\text{20}}{\text{7}}\text{,}\frac{\text{80}}{\text{49}}\text{,}...\\ 4.\frac{-\text{3}}{\text{4}}\text{,}\frac{\text{3}}{\text{16}}\text{,}\frac{\text{-3}}{\text{64}}\text{,}...\\ {\text{5. Prove that:3}}^{\frac{\text{1}}{\text{2}}}{\text{×3}}^{\frac{\text{1}}{\text{4}}}{\text{×3}}^{\frac{\text{1}}{\text{8}}}...\text{=3}\\ {\text{6. Let x = 1+a+a}}^{\text{2}}\text{+}...{\text{and y = 1+b+b}}^{\text{2}}\text{+}...\text{, where}\left|\text{a}\right|\text{<1 and}\left|\text{b}\right|\text{>1.}\\ {\text{Prove that 1+ab+a}}^{\text{2}}{\text{b}}^{\text{2}}\text{+}...\text{=}\frac{\text{xy}}{\text{x+y-1}}\end{array}$

Ans

$\begin{array}{l}1.\text{\hspace{0.17em}Given G.P. is:}\\ \text{1,}\frac{1}{3},\frac{1}{9},...\\ \mathrm{a}=1,\text{r}=\frac{\left(\frac{1}{3}\right)}{1}=\frac{1}{3}\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\\ =\frac{1}{1-\frac{1}{3}}\\ =\frac{1}{\frac{2}{3}}\\ =\frac{3}{2}=1.5\\ \text{2. The given G.P. is:}\\ \text{6},\text{1}.\text{2},0.\text{24},\dots \\ \mathrm{a}=6,\\ \text{r}=\frac{1.2}{6}=0.2\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\\ =\frac{6}{1-0.2}\\ =\frac{6}{0.8}\\ =\frac{60}{8}\\ =7.5\\ \mathrm{Thus},\text{the sum of infinite terms is 7.5.}\\ \text{3. The given G.P. is:}\\ 5,\frac{20}{7},\frac{80}{49},...\\ \mathrm{a}=5,\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}=\frac{\left(\frac{20}{7}\right)}{5}=\frac{4}{7}\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\\ =\frac{5}{1-\frac{4}{7}}\\ =\frac{5}{\left(\frac{3}{7}\right)}\\ =\frac{35}{3}\\ =11.6\\ \mathrm{Thus},\text{the sum of infinite terms is 11.6.}\\ \text{4. The given G.P. is:}\\ \frac{-3}{4},\frac{3}{16},\frac{-3}{64},...\\ \mathrm{a}=\frac{-3}{4},\\ \text{\hspace{0.17em}r}=\frac{\left(\frac{3}{16}\right)}{\left(\frac{-3}{4}\right)}\\ =-\frac{1}{4}\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\left(-\frac{3}{4}\right)}{1-\left(\frac{-1}{4}\right)}=-\frac{3}{4}\times \frac{4}{5}=-\frac{3}{5}\\ \text{5. L.H.S.=}{3}^{\frac{1}{2}}\times 3^{\frac{1}{4}}\times 3^{\frac{1}{8}}...\\ =3^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}\\ =3^{\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)}\left[âˆ\mu \mathrm{S}\infty =\frac{\mathrm{a}}{1-\mathrm{r}}\right]\\ =3^{\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)}\\ =3^1\\ =3=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{, it was to be proved.}\\ \text{6. x =}1+\mathrm{a}+{\mathrm{a}}^2+...\\ =\frac{1}{1-\mathrm{a}}\left[{\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\right]\\ \text{and y}=1+\mathrm{b}+{\mathrm{b}}^2+...\\ =\frac{1}{1-\mathrm{b}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=1+\mathrm{ab}+{\mathrm{a}}^2{\mathrm{b}}^2+...\\ =\frac{1}{1-\mathrm{ab}}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}-1}\\ =\frac{\left(\frac{1}{1-\mathrm{a}}\right)\left(\frac{1}{1-\mathrm{b}}\right)}{\left(\frac{1}{1-\mathrm{a}}\right)+\left(\frac{1}{1-\mathrm{b}}\right)-1}\\ =\frac{1}{1-\mathrm{b}+1-\mathrm{a}-\left(1-\mathrm{a}\right)\left(1-\mathrm{b}\right)}\\ =\frac{1}{2-\mathrm{b}-\mathrm{a}-\left(1-\mathrm{b}-\mathrm{a}+\mathrm{ab}\right)}\\ =\frac{1}{2-\mathrm{b}-\mathrm{a}-1+\mathrm{b}+\mathrm{a}-\mathrm{ab}}\\ =\frac{1}{1-\mathrm{ab}}\\ \text{So,L.H.S.=R.H.S.}\end{array}$

Q.59 Show that the sum of (m + n)^th and (m – n)^th terms of an A.P. is equal to twice the m^th term.

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be a and common difference be d.}\\ \text{Then,}\\ {\text{T}}_{\text{m+n}}+{\text{T}}_{\text{m-n}}=\{\mathrm{a}+(\mathrm{m}+\mathrm{n}-1)\mathrm{d}\}+\{\mathrm{a}+(\mathrm{m}-\mathrm{n}-1)\mathrm{d}\}\\ =2\mathrm{a}+(\mathrm{m}+\mathrm{n}-1+\mathrm{m}-\mathrm{n}-1)\mathrm{d}\\ =2\mathrm{a}+2(\mathrm{m}-1)\mathrm{d}\\ =2\{\mathrm{a}+(\mathrm{m}-1)\mathrm{d}\}\\ =2{\mathrm{T}}_{\mathrm{m}}\\ \mathrm{Thus},\\ {\text{T}}_{\text{m+n}}+{\text{T}}_{\text{m-n}}=2{\mathrm{T}}_{\mathrm{m}}\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.60 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Ans

$\begin{array}{l}\mathrm{Let}\text{three numbers in A.P. are a}-\mathrm{d},\mathrm{a},\mathrm{a}+\mathrm{d}.\\ \mathrm{Then},\text{according to first condition:}\end{array}$ $\begin{array}{l}(\text{a}-\mathrm{d})+\mathrm{a}+(\mathrm{a}+\mathrm{d})=24\\ \Rightarrow 3\mathrm{a}=24\\ \Rightarrow \mathrm{a}=\frac{24}{3}\\ =8\\ \mathrm{According}\text{to second condition:}\\ (\text{a}-\mathrm{d})\mathrm{a}(\mathrm{a}+\mathrm{d})=440\\ \Rightarrow (\text{8}-\mathrm{d})8(8+\mathrm{d})=440\\ \Rightarrow (\text{8}-\mathrm{d})(8+\mathrm{d})=\frac{440}{8}\\ \Rightarrow 64-{\mathrm{d}}^2=55\\ \Rightarrow {\mathrm{d}}^2=64-55\\ =9\\ \Rightarrow \mathrm{d}=\pm 3\\ \mathrm{The}\text{three numbers in A.P. are:}\\ \text{8}-3,8,8+3\text{or 8}+3,8,8-3\\ \Rightarrow 5,8,11\text{or 11,\hspace{0.17em}8,\hspace{0.17em}5}\\ \text{Thus, the three numbers are 5, 8, 11.}\end{array}$

Q.61 Let the sum of n, 2n, 3n terms of an A.P. be S₁, S₂ and S₃, respectively, show that S₃ = 3(S₂ – S₁).

Ans

$\begin{array}{l}\begin{array}{l}âˆ\mu {\text{S}}_1={\text{S}}_{\text{n}}\\ =\frac{\text{n}}{2}\left\{2\text{a}+\left(\text{n}-1\right)\text{d}\right\}\\ {\text{S}}_2={\text{S}}_{2\text{n}}\end{array}\\ \begin{array}{l}=\frac{2\text{n}}{2}\left\{2\text{a}+\left(2\text{n}-1\right)\text{d}\right\}\\ {\text{S}}_3={\text{S}}_{3\text{n}}\\ =\frac{3\text{n}}{2}\left\{2\text{a}+\left(3\text{n}-1\right)\text{d}\right\}\end{array}\\ \begin{array}{l}\text{R}.\text{H}.\text{S}.=\text{3}\left({\text{S}}_{\text{2}}-{\text{S}}_{\text{1}}\right)\\ =\text{3}\left[\frac{2\text{n}}{2}\left\{2\text{a}+\left(2\text{n}-1\right)\text{d}\right\}-\frac{\text{n}}{2}\left\{2\text{a}+\left(\text{n}-1\right)\text{d}\right\}\right]\end{array}\\ =\frac{3\text{n}}{2}\left[2\left\{2\text{a}+\left(2\text{n}-1\right)\text{d}\right\}-\left\{2\text{a}+\left(\text{n}-1\right)\text{d}\right\}\right]\\ =\frac{3\text{n}}{2}\left[4\text{a}-2\text{a}+\left(4\text{n}-2-\text{n}+1\right)\text{d}\right]\\ =\frac{3\text{n}}{2}\left\{2\text{a}+\left(3\text{n}-1\right)\text{d}\right\}\\ \begin{array}{l}={\text{S}}_3=\text{L}.\text{H}.\text{S}.\\ \mathrm{Thus},{\text{S}}_3=\text{3}\left({\text{S}}_{\text{2}}-{\text{S}}_{\text{1}}\right)\text{is proved.}\end{array}\end{array}$

Q.62 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Ans

$\begin{array}{l}\mathrm{The}\text{numbers between 200 to 400 divisible by 7 are:}\\ 203,210,217,224,231,...,399.\\ \mathrm{Here},\text{a}=203,\text{d}=210-203=7,\mathrm{l}=399\\ âˆ\mu \mathrm{l}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 399=203+(\mathrm{n}-1)7\\ \Rightarrow \frac{196}{7}=\mathrm{n}-1\end{array}$ $\begin{array}{l}\Rightarrow 28+1=\mathrm{n}\\ \Rightarrow \mathrm{n}=29\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})\\ \Rightarrow {\mathrm{S}}_{29}=\frac{29}{2}(203+399)\\ =\frac{29}{2}\times 602\\ =8729\\ \mathrm{Thus},\text{the sum of multiples of 7 between 200}\\ \text{and 400 is 8729.}\end{array}$

Q.63 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Ans

$\begin{array}{l}\mathrm{The}\text{numbers from 1 to 100 divisible by 2:}\\ 2,4,6,8,...,100\\ \mathrm{Here},\text{a}=2,\text{d}=4-2=2,\mathrm{l}=100\\ âˆ\mu \mathrm{l}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 100=2+(\mathrm{n}-1)2\\ \frac{98}{2}=\mathrm{n}-1\\ \Rightarrow \mathrm{n}=49+1\\ =50\\ âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})\end{array}$ $\begin{array}{l}\therefore {\mathrm{S}}_{50}=\frac{50}{2}(2+100)\\ =25\times 102\\ =2550\\ \mathrm{The}\text{numbers from 1 to 100 divisible by 5:}\\ 5,10,15,20,...,100\\ \mathrm{Here},\text{a}=5,\text{d}=10-5=5,\mathrm{l}=100\\ âˆ\mu \mathrm{l}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 100=5+(\mathrm{n}-1)5\\ \frac{95}{5}=\mathrm{n}-1\\ \Rightarrow \mathrm{n}=19+1\\ =20\\ âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})\\ \therefore {\mathrm{S}}_{20}=\frac{20}{2}(5+100)\\ =10\times 105\\ =1050\\ \mathrm{The}\text{numbers from 1 to 100 divisible by 2 and 5:}\\ \text{10,20,30,}...\text{,100}\\ \mathrm{Here},\text{a}=10,\text{d}=20-10=10,\mathrm{l}=100\\ âˆ\mu \mathrm{l}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 100=10+(\mathrm{n}-1)10\\ \Rightarrow \frac{90}{10}=\mathrm{n}-1\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow \mathrm{n}=10\\ âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})\\ \therefore {\mathrm{S}}_{10}=\frac{10}{2}(10+100)\\ =5\times 110\\ =550\\ \mathrm{Sum}\text{of numbers divisible by 2 or 5}\\ =\mathrm{Sum}\text{of multiples of 2}+\mathrm{Sum}\text{of multiples of 5}\\ -\mathrm{Sum}\text{of multiples of 10}\\ =2550+1050-550\\ =3050\end{array}$

Q.64 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Ans

$\begin{array}{l}\mathrm{Two}\text{digit numbers which when divided by 4, yields}\\ \text{1 as remainder are:}\\ 13,17,21,...,97\\ \mathrm{a}=13,\mathrm{d}=17-13=4,\mathrm{l}=97\\ âˆ\mu \mathrm{l}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 97=13+(\mathrm{n}-1)\mathrm{d}\\ 97-13=(\mathrm{n}-1)4\\ \frac{84}{4}=\mathrm{n}-1\\ \mathrm{n}=21+1\\ =22\end{array}$ $\begin{array}{l}âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})\\ \therefore {\mathrm{S}}_{22}=\frac{22}{2}(13+97)\\ =\frac{22}{2}\times 110\\ =1210\\ \mathrm{Thus},\text{the sum of 22 terms is 1210.}\end{array}$

Q.65

$\begin{array}{l}\text{If f is a function satisfying f}\left(\text{x+y}\right)\text{=f}\left(\text{x}\right)\text{f}\left(\text{y}\right)\text{for all x, y}\in \text{N such that f}\left(\text{1}\right)\text{=3 and}\\ \sum _{\text{x=1}}^{\text{n}}\text{f}\left(\text{x}\right)\text{=120, find the value of n.}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)\text{, f}\left(1\right)=3\text{\hspace{0.17em}and}\sum _{\mathrm{x}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{x}\right)=120.\\ \mathrm{So},\mathrm{f}\left(2\right)=\mathrm{f}(1+1)\\ =\mathrm{f}\left(1\right)\mathrm{f}\left(1\right)\\ =3\times 3\\ =3^2\\ \mathrm{f}\left(3\right)=\mathrm{f}(2+1)\\ =\mathrm{f}\left(2\right)\mathrm{f}\left(1\right)\\ =3^2\times 3\\ =3^3\\ \mathrm{f}\left(4\right)=\mathrm{f}(2+2)\\ =\mathrm{f}\left(2\right)\mathrm{f}\left(2\right)\end{array}$ $\begin{array}{l}=3^2\times 3^2\\ =3^4\\ âˆ\mu \sum _{\mathrm{x}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{x}\right)=120\\ \Rightarrow \mathrm{f}\left(1\right)+\mathrm{f}\left(2\right)+\mathrm{f}\left(3\right)+\mathrm{f}\left(4\right)+...+\mathrm{f}\left(\mathrm{n}\right)=120\\ 3+3^2+3^3+3^4+...+3^{\mathrm{n}}=120\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow \frac{3(3^{\mathrm{n}}-1)}{3-1}=120\\ \Rightarrow 3^{\mathrm{n}}-1=120\times \frac{2}{3}\\ \Rightarrow 3^{\mathrm{n}}=80+1\\ \Rightarrow 3^{\mathrm{n}}=3^4\\ \Rightarrow \mathrm{n}=4\\ \mathrm{Thus},\text{the value of n is 4.}\end{array}$

Q.66 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\text{5}\\ \mathrm{Common}\text{difference of G.P.}=\text{2}\\ \text{Let number of terms in G.P.}=\text{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sum of n terms in G.P.}=315\\ 5\times \frac{(2^{\mathrm{n}}-1)}{2-1}=315\end{array}$ $\begin{array}{l}\Rightarrow 2^{\mathrm{n}}-1=\frac{315}{5}\\ 2^{\mathrm{n}}=63+1\\ \Rightarrow 2^{\mathrm{n}}=2^6\\ \Rightarrow \mathrm{n}=6\\ âˆ\mu \mathrm{l}={\mathrm{ar}}^{\mathrm{n}-1}\\ =5\times 2^{(6-1)}\\ =5\times 32\\ =160\\ \mathrm{Thus},\text{the last term of G.P. is 160 and number of}\\ \text{terms in G.P. is 6.}\end{array}$

Q.67 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=1\\ \mathrm{Let}\text{common ratio}=\mathrm{r}\\ {\mathrm{T}}_3+{\mathrm{T}}_5=90\\ \Rightarrow {\mathrm{ar}}^2+{\mathrm{ar}}^4=90\\ \Rightarrow 1.{\mathrm{r}}^2+1.{\mathrm{r}}^4=90\\ \Rightarrow {\mathrm{r}}^4+{\mathrm{r}}^2=90\\ \Rightarrow {\mathrm{r}}^4+{\mathrm{r}}^2-90=0\\ \Rightarrow ({\mathrm{r}}^2+10)({\mathrm{r}}^2-9)=0\\ \Rightarrow {\mathrm{r}}^2=9,-10\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{r}=\pm 3[\mathrm{Neglecting}{\text{r}}^{\text{2}}=-10]\\ \mathrm{Thus},\text{the common ratio of G.P. is}\pm 3.\end{array}$

Q.68 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Ans

$\begin{array}{l}\mathrm{Let}\text{three terms in G.P. are}\frac{\mathrm{a}}{\mathrm{r}},\mathrm{a},\mathrm{ar}.\\ \mathrm{Then},\mathrm{according}\mathrm{to}\text{first condition, we have}\\ \frac{\mathrm{a}}{\mathrm{r}}+\mathrm{a}+\mathrm{ar}=56...\left(\mathrm{i}\right)\\ \text{On subtracting 1},\text{7},\text{21 frome each number respectively,}\\ (\frac{\text{a}}{\mathrm{r}}-\text{1}),(\mathrm{a}-7),(\mathrm{ar}-21)\text{are in A.P., then}\\ 2(\mathrm{a}-7)=(\frac{\text{a}}{\mathrm{r}}-\text{1})+(\mathrm{ar}-21)\\ 2\mathrm{a}-14=\frac{\text{a}}{\mathrm{r}}+\mathrm{ar}-22\\ \Rightarrow \frac{\text{a}}{\mathrm{r}}+\mathrm{ar}-2\mathrm{a}=8...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}56}-\mathrm{a}-2\mathrm{a}=8\end{array}$ $\begin{array}{l}\Rightarrow 56-8=3\mathrm{a}\\ \Rightarrow \mathrm{a}=\frac{48}{3}=16\\ \mathrm{Putting}\text{a}=\text{14 in equation}\left(\mathrm{i}\right),\mathrm{we}\text{get}\\ \frac{16}{\mathrm{r}}+16+16\mathrm{r}=56\\ \Rightarrow \frac{1}{\mathrm{r}}+\mathrm{r}=\frac{56-16}{16}\\ \Rightarrow \frac{1}{\mathrm{r}}+\mathrm{r}=\frac{40}{16}\\ \Rightarrow 2+2{\mathrm{r}}^2=5\mathrm{r}\\ \Rightarrow 2{\mathrm{r}}^2-5\mathrm{r}+2=0\\ \Rightarrow 2{\mathrm{r}}^2-4\mathrm{r}-\mathrm{r}+2=0\\ \Rightarrow 2\mathrm{r}(\mathrm{r}-2)-1(\mathrm{r}-2)=0\\ \Rightarrow (\mathrm{r}-2)(2\mathrm{r}-1)=0\\ \Rightarrow \mathrm{r}=2,\frac{1}{2}\\ \mathrm{So},\text{three numbers when a}=16\text{and r}=2,\frac{1}{2},\\ 16\times 2,16,\frac{16}{2}\text{or}\frac{\text{16}}{2}\text{,16,16}\times \text{2}\\ \mathrm{i}.\mathrm{e}.,\text{32,16,8 or 8,16,32}\\ \text{Thus, the required three numbers are 8,16,32.}\end{array}$

Q.69 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Ans

$\begin{array}{l}\mathrm{Let}\text{G.P. containing even number of terms is:}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{,}...{\text{, ar}}^{\text{n}},{\text{ar}}^{\text{n+1}},.\dots ,{\mathrm{ar}}^{2\mathrm{n}-1},{\mathrm{ar}}^{2\mathrm{n}}\\ \mathrm{According}\text{to given condition,}\\ {\mathrm{S}}_{2\mathrm{n}}=5\times (\mathrm{a}+{\mathrm{ar}}^2+{\mathrm{ar}}^4+...+{\mathrm{ar}}^{2\mathrm{n}-1})\\ \left(\begin{array}{l}\text{a}+\text{ar}+{\text{ar}}^{\text{2}}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}}\\ +{\text{ar}}^{\text{n+1}}+...+{\mathrm{ar}}^{2\mathrm{n}-1}+{\mathrm{ar}}^{2\mathrm{n}}\end{array}\right)=5\times (\mathrm{a}+{\mathrm{ar}}^2+{\mathrm{ar}}^4+...+{\mathrm{ar}}^{2\mathrm{n}-1})\end{array}$ $\begin{array}{l}(\text{ar}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}}+...+{\mathrm{ar}}^{2\mathrm{n}})=4\times (\mathrm{a}+{\mathrm{ar}}^2+{\mathrm{ar}}^4+...+{\mathrm{ar}}^{2\mathrm{n}-1})\\ \frac{\mathrm{ar}({\mathrm{r}}^{2\mathrm{n}}-1)}{\mathrm{r}-1}=4\times \frac{\mathrm{a}\{{\left({\mathrm{r}}^2\right)}^{\mathrm{n}}-1\}}{\mathrm{r}-1}\\ \Rightarrow \mathrm{r}({\mathrm{r}}^{2\mathrm{n}}-1)=4({\mathrm{r}}^{2\mathrm{n}}-1)\\ \Rightarrow \mathrm{r}=4\\ \mathrm{Thus},\text{the common ratio of G.P. is 4.}\end{array}$

Q.70 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Ans

$\begin{array}{l}\mathrm{Let}\text{common difference of an A.P.}=\text{d}\\ \text{Let number of terms in an A}.\text{P}.=\mathrm{n}\\ \mathrm{The}\text{first term of A.P.}=\text{11}\\ \text{Sum of the first four terms of an A}.\text{P}.=56\end{array}$ $\begin{array}{l}\frac{4}{2}\{2\times 11+(4-1)\mathrm{d}\}=56[\mathrm{Sn}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}]\\ 3\mathrm{d}=28-22\\ \mathrm{d}=\frac{6}{3}\\ =2\\ \text{Sum of the last four terms\hspace{0.17em}of an A}.\text{P}.=112\\ \{\mathrm{a}+(\mathrm{n}-4)\mathrm{d}\}+\{\mathrm{a}+(\mathrm{n}-3)\mathrm{d}\}+\{\mathrm{a}+(\mathrm{n}-2)\mathrm{d}\}\\ +\{\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}=112\\ \{11+(\mathrm{n}-4)2\}+\{11+(\mathrm{n}-3)2\}+\{11+(\mathrm{n}-2)2\}\\ +\{11+(\mathrm{n}-1)2\}=112\\ 44+(\mathrm{n}-4+\mathrm{n}-3+\mathrm{n}-2+\mathrm{n}-1)2=112\\ 4\mathrm{n}-10=\frac{112-44}{2}\\ \mathrm{n}=\frac{34+10}{4}\\ =11\\ \mathrm{Thus},\text{the number of terms in an A.P. is 11.}\end{array}$

Q.71

$\mathbf{If}\mathbf{}\frac{\mathbf{a}\mathbf{+}\mathbf{bx}}{\mathbf{a}\mathbf{-}\mathbf{bx}}\mathbf{=}\frac{\mathbf{b}\mathbf{+}\mathbf{cx}}{\mathbf{b}\mathbf{-}\mathbf{cx}}\mathbf{=}\frac{\mathbf{c}\mathbf{+}\mathbf{dx}}{\mathbf{c}\mathbf{-}\mathbf{dx}}(x\ne 0)\mathbf{,}\mathbf{}\mathbf{\text{then show that a, b, c and d are in G.P}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given:}\\ \frac{\mathrm{a}+\mathrm{bx}}{\mathrm{a}-\mathrm{bx}}=\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}=\frac{\mathrm{c}+\mathrm{dx}}{\mathrm{c}-\mathrm{dx}}(\mathrm{x}\ne 0)\end{array}$ $\begin{array}{l}\Rightarrow \frac{\mathrm{a}+\mathrm{bx}}{\mathrm{a}-\mathrm{bx}}=\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}\\ \Rightarrow (\mathrm{a}+\mathrm{bx})(\mathrm{b}-\mathrm{cx})=(\mathrm{b}+\mathrm{cx})(\mathrm{a}-\mathrm{bx})\\ \Rightarrow \mathrm{ab}-\mathrm{acx}+{\mathrm{b}}^2\mathrm{x}-{\mathrm{bcx}}^2=\mathrm{ba}-{\mathrm{b}}^2\mathrm{x}+\mathrm{cax}-{\mathrm{cbx}}^2\\ \Rightarrow 2{\mathrm{b}}^2\mathrm{x}=2\mathrm{acx}\\ \Rightarrow {\mathrm{b}}^2=\mathrm{ac}\\ \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}...\left(\mathrm{i}\right)\\ \mathrm{And}\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}=\frac{\mathrm{c}+\mathrm{dx}}{\mathrm{c}-\mathrm{dx}}\\ \Rightarrow (\mathrm{b}+\mathrm{cx})(\mathrm{c}-\mathrm{dx})=(\mathrm{c}+\mathrm{dx})(\mathrm{b}-\mathrm{cx})\\ \Rightarrow \mathrm{bc}-\mathrm{bdx}+{\mathrm{c}}^2\mathrm{x}-{\mathrm{cdx}}^2=\mathrm{cb}-{\mathrm{c}}^2\mathrm{x}+\mathrm{dbx}-{\mathrm{dcx}}^2\\ \Rightarrow 2{\mathrm{c}}^2\mathrm{x}=2\mathrm{bdx}\\ \Rightarrow {\mathrm{c}}^2=\mathrm{bd}\\ \Rightarrow \frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}\\ \Rightarrow \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{are in G.P.}\end{array}$

Q.72 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ.

Ans

$\begin{array}{l}\mathrm{Let}\text{n terms of G.P. are}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{,}...{\text{, ar}}^{\text{n}-3},{\text{ar}}^{\text{n}-2},{\text{ar}}^{\text{n}-1}\end{array}$ $\begin{array}{l}\mathrm{Then}\text{according to the given conditions:}\\ \text{S}=\text{a}+\text{ar}+{\text{ar}}^{\text{2}}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}-3}+{\text{ar}}^{\text{n}-2}+{\text{ar}}^{\text{n}-1}\\ =\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{(\mathrm{r}-1)}\\ \mathrm{P}=\text{a}\times \text{ar}\times {\text{ar}}^{\text{2}}\times {\text{ar}}^{\text{3}}\times ...\times {\text{ar}}^{\text{n}-3}\times {\text{ar}}^{\text{n}-2}\times {\text{ar}}^{\text{n}-1}\\ ={\mathrm{a}}^{\mathrm{n}}\times {\mathrm{r}}^{1+2+3+...+(\mathrm{n}-3)+(\mathrm{n}-2)+(\mathrm{n}-1)}\\ ={\mathrm{a}}^{\mathrm{n}}\times {\mathrm{r}}^{\frac{(\mathrm{n}-1)(1+\mathrm{n}-1)}{2}}\\ ={\mathrm{a}}^{\mathrm{n}}\times {\mathrm{r}}^{\frac{\mathrm{n}(\mathrm{n}-1)}{2}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {\text{P}}^{\text{2}}={\mathrm{a}}^{2\mathrm{n}}\times {\mathrm{r}}^{\mathrm{n}(\mathrm{n}-1)}\\ \mathrm{R}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{ar}}+\frac{1}{{\mathrm{ar}}^2}+\frac{1}{{\mathrm{ar}}^3}+...+\frac{1}{{\mathrm{ar}}^{\mathrm{n}-1}}\\ =\frac{\frac{1}{\mathrm{a}}(1-\frac{1}{{\mathrm{r}}^{\mathrm{n}}})}{(1-\frac{1}{\mathrm{r}})}\\ =\frac{\mathrm{r}}{{\mathrm{ar}}^{\mathrm{n}}}\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)\\ \therefore {\mathrm{R}}^{\mathrm{n}}=\frac{{\mathrm{r}}^{\mathrm{n}(1-\mathrm{n})}}{{\mathrm{a}}^{\mathrm{n}}}{\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{P}}^2{\mathrm{R}}^{\mathrm{n}}\\ ={\mathrm{a}}^{2\mathrm{n}}\times {\mathrm{r}}^{\mathrm{n}(\mathrm{n}-1)}\times \frac{{\mathrm{r}}^{\mathrm{n}(1-\mathrm{n})}}{{\mathrm{a}}^{\mathrm{n}}}{\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\\ ={\mathrm{a}}^{\mathrm{n}}\times {{\mathrm{r}}^{{\mathrm{n}}^2-\mathrm{n}}}^{+\mathrm{n}-{\mathrm{n}}^2}\times {\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}={\mathrm{a}}^{\mathrm{n}}\times {\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\\ ={\left\{\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}\right\}}^{\mathrm{n}}\\ ={\mathrm{S}}^{\mathrm{n}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.73 The p^th, q^th and r^th terms of an A.P. are a, b, c, respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0.

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. is A and common difference is D.}\\ \text{Then, according to given condition:}\\ {\mathrm{T}}_{\mathrm{p}}=\mathrm{a}\\ \Rightarrow \mathrm{A}+(\mathrm{p}-1)\mathrm{D}=\mathrm{a}...\left(\mathrm{i}\right)\\ {\mathrm{T}}_{\mathrm{q}}=\mathrm{b}\\ \Rightarrow \mathrm{A}+(\mathrm{q}-1)\mathrm{D}=\mathrm{b}...\left(\mathrm{ii}\right)\\ {\mathrm{T}}_{\mathrm{r}}=\mathrm{c}\\ \Rightarrow \mathrm{A}+(\mathrm{r}-1)\mathrm{D}=\mathrm{c}...\left(\mathrm{iii}\right)\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=(\text{q}-\text{r})\text{a}+(\text{r}-\text{p})\text{b}+(\text{p}-\text{q})\text{c}\\ =(\text{q}-\text{r})\{\mathrm{A}+(\mathrm{p}-1)\mathrm{D}\}+(\text{r}-\text{p})\{\mathrm{A}+(\mathrm{q}-1)\mathrm{D}\}\\ +(\text{p}-\text{q})\{\mathrm{A}+(\mathrm{r}-1)\mathrm{D}\}\\ =\mathrm{A}(\text{q}-\text{r}+\text{r}-\text{p}+\text{p}-\text{q})+\mathrm{D}\left\{\begin{array}{l}(\text{q}-\text{r})(\mathrm{p}-1)+(\text{r}-\text{p})(\mathrm{q}-1)\\ +(\text{p}-\text{q})(\mathrm{r}-1)\end{array}\right\}\\ =\mathrm{A}\left(0\right)+\mathrm{D}(\mathrm{qp}-\mathrm{q}-\mathrm{rp}+\mathrm{r}+\mathrm{rq}-\mathrm{r}-\mathrm{pq}+\mathrm{p}+\mathrm{pr}-\mathrm{p}-\mathrm{qr}+\mathrm{q})\\ =0+0\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.74

$\mathbf{\text{If a}}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\mathbf{\text{,\hspace{0.17em}b}}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\mathbf{\text{,\hspace{0.17em}c}}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\mathbf{\text{are in A.P., prove that a, b and c are in A.P.}}$

Ans

$\begin{array}{l}\text{Since,a}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\text{,\hspace{0.17em}b}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\text{,c}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\text{areinA.P.}\\ \Rightarrow \left(\frac{\text{a}}{\text{b}}\text{+}\frac{\text{a}}{\text{c}}\right)\text{,}\left(\frac{\text{b}}{\text{c}}\text{+}\frac{\text{b}}{\text{a}}\right)\text{,\hspace{0.17em}}\left(\frac{\text{c}}{\text{a}}\text{+}\frac{\text{c}}{\text{b}}\right)\text{areinA.P.}\\ \Rightarrow \left(\frac{\text{a}}{\text{a}}\text{+}\frac{\text{a}}{\text{b}}\text{+}\frac{\text{a}}{\text{c}}\right)\text{,}\left(\frac{\text{b}}{\text{b}}\text{+}\frac{\text{b}}{\text{c}}\text{+}\frac{\text{b}}{\text{a}}\right)\text{,\hspace{0.17em}}\left(\frac{\text{c}}{\text{c}}\text{+}\frac{\text{c}}{\text{a}}\text{+}\frac{\text{c}}{\text{b}}\right)\text{areinA.P.}\\ \Rightarrow \text{a}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\text{,b}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\text{,\hspace{0.17em}c}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\text{areinA.P.}\\ \mathrm{\Rightarrow }\text{a,b,careinA.P.}\end{array}$

Q.75 If a, b, c, d are in G.P, prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

Ans

$\begin{array}{l}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{are in G}.\text{P}\\ \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ \Rightarrow \mathrm{b}=\mathrm{ar},\mathrm{c}=\mathrm{br}\text{and d}=\text{cr}\\ \Rightarrow \mathrm{b}=\mathrm{ar},\mathrm{c}={\mathrm{ar}}^2\text{and d}={\text{ar}}^{\text{3}}\\ \mathrm{Then},\\ \frac{({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}})}{({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}})}=\frac{({\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}})}{({\text{a}}^{\text{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}})}\\ =\frac{{\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}(1+{\mathrm{r}}^{\mathrm{n}})}{{\text{a}}^{\text{n}}(\text{1}+{\mathrm{r}}^{\mathrm{n}})}\\ ={\mathrm{r}}^{\mathrm{n}}\\ \frac{({\text{c}}^{\text{n}}+{\text{d}}^{\text{n}})}{({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}})}=\frac{({\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{3\mathrm{n}})}{({\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}})}\\ =\frac{{\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}}(1+{\mathrm{r}}^{\mathrm{n}})}{{\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}(\text{1}+{\mathrm{r}}^{\mathrm{n}})}\\ ={\mathrm{r}}^{\mathrm{n}}\\ \mathrm{So},\frac{({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}})}{({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}})}=\frac{({\text{c}}^{\text{n}}+{\text{d}}^{\text{n}})}{({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}})}\\ \mathrm{Therefore},({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}),({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}}),({\text{c}}^{\text{n}}+{\text{d}}^{\text{n}})\text{are G.P.}\end{array}$

Q.76 If a and b are the roots of x² – 3x + p = 0 and c, d are roots of x² – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

Ans

$\begin{array}{l}\mathrm{a}{\text{and b are the roots of x}}^{\text{2}}-\text{3x}+\text{p}=0,\text{then}\\ \text{\hspace{0.17em}sum of zeroes}=-\frac{-3}{1}\\ \mathrm{a}+\mathrm{b}=3...\left(\mathrm{i}\right)\\ \text{Product of zeroes}\\ =\frac{\mathrm{p}}{1}\\ \mathrm{ab}=\frac{\mathrm{p}}{1}\\ \Rightarrow \mathrm{ab}=\mathrm{p}...\left(\mathrm{ii}\right)\\ \mathrm{c}{\text{and d are the roots of x}}^{\text{2}}-\text{12x}+\text{q}=0,\text{then}\\ \text{sum of zeroes}=-\frac{-12}{1}\\ \mathrm{c}+\mathrm{d}=12\dots \left(\mathrm{iii}\right)\\ \text{Product of zeroes}\\ =\frac{\mathrm{q}}{1}\\ \mathrm{cd}=\frac{\mathrm{q}}{1}\\ \Rightarrow \mathrm{cd}=\mathrm{q}...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{a,b,c,d are in G.P.}\\ \text{So,}\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ \Rightarrow \mathrm{b}=\mathrm{ar},\text{c}=\text{br, d}=\text{cr}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{b}=\mathrm{ar},\text{c}={\text{ar}}^{\text{2}}\text{, d}={\text{ar}}^{\text{3}}\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{a}+\mathrm{ar}=3\\ \mathrm{a}(1+\mathrm{r})=3...\left(\mathrm{v}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{iii}\right),\text{we get}\\ {\mathrm{ar}}^2+{\mathrm{ar}}^3=12\\ {\mathrm{ar}}^2(1+\mathrm{r})=12...\left(\mathrm{vi}\right)\\ \mathrm{Dividing}\text{equation}\left(\mathrm{vi}\right)\text{by equation}\left(\mathrm{v}\right),\text{\hspace{0.17em}we get}\\ \frac{{\mathrm{ar}}^2(1+\mathrm{r})}{\mathrm{a}(1+\mathrm{r})}=\frac{12}{3}\\ \Rightarrow {\mathrm{r}}^2=4\\ \Rightarrow \mathrm{r}=\pm 2\\ \mathrm{Putting}\text{r}=\text{2, in equation}\left(\mathrm{v}\right),\mathrm{we}\mathrm{get}\\ \mathrm{a}(1+2)=3\\ \Rightarrow \mathrm{a}=\frac{3}{3}=1\\ \mathrm{Putting}\text{r}=-\text{2, in equation}\left(\mathrm{v}\right),\mathrm{we}\mathrm{get}\\ \mathrm{a}(1-2)=3\\ \Rightarrow \mathrm{a}=\frac{3}{-1}=-3\\ \mathrm{Case}\text{I:}\mathrm{If}\text{a}=\text{1 and r}=\text{2, then}\\ \text{a}=1,\mathrm{b}=1\times 2,\text{c}=1\times 2^{\text{2}}\text{, d}=1\times 2^{\text{3}}\\ \Rightarrow \mathrm{a}=1,\mathrm{b}=2,\mathrm{c}=4,\mathrm{d}=8\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{q}+\mathrm{p}}{\mathrm{q}-\mathrm{p}}\\ =\frac{\mathrm{cd}+\mathrm{ab}}{\mathrm{cd}-\mathrm{ab}}\\ =\frac{4\times 8+1\times 2}{4\times 8-1\times 2}\\ =\frac{32+2}{32-2}\\ =\frac{34}{30}\\ =\frac{17}{15}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Case}\text{II:}\mathrm{If}\text{a}=-3\text{and r}=-\text{2, then}\\ \text{a}=-3,\mathrm{b}=-3\times -2,\text{c}=-3\times {(-2)}^{\text{2}}\text{, d}=-3\times {(-2)}^{\text{3}}\\ \Rightarrow \mathrm{a}=-3,\mathrm{b}=6,\mathrm{c}=-12,\mathrm{d}=24\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{q}+\mathrm{p}}{\mathrm{q}-\mathrm{p}}\\ =\frac{\mathrm{cd}+\mathrm{ab}}{\mathrm{cd}-\mathrm{ab}}\\ =\frac{-12\times 24+(-3)\times 6}{-12\times 24-(-3)\times 6}\\ =\frac{-288-18}{-288+18}\\ =\frac{-306}{-270}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{17}{15}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},(\mathrm{q}+\mathrm{p}):(\mathrm{q}-\mathrm{p})=17:15.\mathrm{Hence}\text{proved.}\end{array}$

Q.77

$\begin{array}{l}\text{The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n.}\\ \text{Show that a : b =}\left(\text{m +}\sqrt{{\text{m}}^{\text{2}}-{\text{n}}^{\text{2}}}\right)\text{:}\left(\text{m}-\sqrt{{\text{m}}^{\text{2}}-{\text{n}}^{\text{2}}}\right)\text{.}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given,}\\ \frac{\mathrm{A}.\mathrm{M}.\text{of a and b}}{\mathrm{G}.\mathrm{M}.\text{of a and b}}=\frac{\mathrm{m}}{\mathrm{n}}\\ \Rightarrow \frac{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}{\sqrt{\mathrm{ab}}}=\frac{\mathrm{m}}{\mathrm{n}}\\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{ab}}}=\frac{\mathrm{m}}{\mathrm{n}}\\ \mathrm{Apply}\text{componendo and dividendo, we get}\\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}+2\sqrt{\mathrm{ab}}}{\mathrm{a}+\mathrm{b}-2\sqrt{\mathrm{ab}}}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}\\ \Rightarrow \frac{{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})}^2}{{(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}})}^2}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}\\ \Rightarrow \frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}}=\frac{\sqrt{\mathrm{m}+\mathrm{n}}}{\sqrt{\mathrm{m}-\mathrm{n}}}\\ \mathrm{Apply}\text{again componendo and dividendo, we get}\\ \frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}-\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}=\frac{\sqrt{\mathrm{m}+\mathrm{n}}+\sqrt{\mathrm{m}-\mathrm{n}}}{\sqrt{\mathrm{m}+\mathrm{n}}-\sqrt{\mathrm{m}-\mathrm{n}}}\end{array}$ $\begin{array}{l}\Rightarrow \frac{2\sqrt{\mathrm{a}}}{2\sqrt{\mathrm{b}}}=\frac{\sqrt{\mathrm{m}+\mathrm{n}}+\sqrt{\mathrm{m}-\mathrm{n}}}{\sqrt{\mathrm{m}+\mathrm{n}}-\sqrt{\mathrm{m}-\mathrm{n}}}\\ \Rightarrow {\left(\frac{\sqrt{\mathrm{a}}}{\sqrt{\mathrm{b}}}\right)}^2=\frac{{(\sqrt{\mathrm{m}+\mathrm{n}}+\sqrt{\mathrm{m}-\mathrm{n}})}^2}{{(\sqrt{\mathrm{m}+\mathrm{n}}-\sqrt{\mathrm{m}-\mathrm{n}})}^2}\left[\mathrm{Squarring}\text{both sides}\right]\\ \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}+2\sqrt{\mathrm{m}+\mathrm{n}}\sqrt{\mathrm{m}-\mathrm{n}}}{\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}-2\sqrt{\mathrm{m}+\mathrm{n}}\sqrt{\mathrm{m}-\mathrm{n}}}\\ \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{2\mathrm{m}+2\sqrt{{\mathrm{m}}^2-{\mathrm{n}}^2}}{2\mathrm{m}-2\sqrt{{\mathrm{m}}^2-{\mathrm{n}}^2}}\\ \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{m}+\sqrt{{\mathrm{m}}^2-{\mathrm{n}}^2}}{\mathrm{m}-\sqrt{{\mathrm{m}}^2-{\mathrm{n}}^2}}\\ \mathrm{Thus},\text{a:b}=(\mathrm{m}+\sqrt{{\mathrm{m}}^2-{\mathrm{n}}^2}):(\mathrm{m}-\sqrt{{\mathrm{m}}^2-{\mathrm{n}}^2}).\end{array}$

Q.78

$\begin{array}{l}\text{If a, b, c are in A.P.; b, c, d are in G.P. and}\frac{\text{1}}{\text{c}}\text{,}\frac{\text{1}}{\text{d}}\text{,}\frac{\text{1}}{\text{e}}\text{are in A.P.}\\ \text{prove that a, c, e are in G.P.}\end{array}$

Ans

$\begin{array}{l}\text{a,b,careinA.P.,then}\\ \text{b=}\frac{\text{a+c}}{\text{2}}\\ 2\mathrm{b}=\mathrm{a}+\mathrm{c}...\left(\text{i}\right)\\ \text{b},\text{c},\text{dareinG.P.,then}\\ {\text{c}}^{\text{2}}=\text{bd}\\ \text{d}=\frac{{\text{c}}^2}{\text{b}}....\left(\mathrm{ii}\right)\\ \frac{1}{\text{c}},\frac{1}{\text{d}},\frac{1}{\text{e}}\mathrm{are}\mathrm{in}\text{A}.\text{P}.,\text{then}\\ \frac{2}{\text{d}}=\frac{1}{\text{c}}+\frac{1}{\text{e}}\\ \frac{2}{\text{d}}=\frac{\text{e}+\text{c}}{\mathrm{ce}}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{andequation}\left(\mathrm{iii}\right),\text{weget}\\ \frac{2}{\left(\frac{{\text{c}}^2}{\text{b}}\right)}=\frac{\text{e}+\text{c}}{\mathrm{ce}}\\ \Rightarrow \frac{2\text{b}}{{\text{c}}^2}=\frac{\text{e}+\text{c}}{\mathrm{ce}}\\ \Rightarrow \frac{2\text{b}}{\text{c}}=\frac{\text{e}+\text{c}}{\text{e}}\\ \Rightarrow \left(\text{a}+\text{c}\right)\text{e}=\mathrm{ce}+{\text{c}}^2\left[\mathrm{From}\text{equation}\left(\text{i}\right)\right]\\ \Rightarrow {\text{ae+ce=ce+c}}^{\text{2}}\\ \Rightarrow {\text{ae=c}}^{\text{2}}\\ \Rightarrow \text{a,c,eareinG.P.}\end{array}$

Q.79 Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + …
(ii) .6 +. 66 +. 666+…

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}{\mathrm{S}}_{\mathrm{n}}=\text{5}+\text{55}+\text{555}+\dots \mathrm{}\\ =5(1+11+111+1111+...)\\ =\frac{5}{9}(9+99+999+9999+...)\\ =\frac{5}{9}\{(10-1)+({10}^2-1)+({10}^3-1)+({10}^4-1)+...\}\\ =\frac{5}{9}\{10+{10}^2+{10}^3+{10}^4+...-1-1-1-1-...\mathrm{n}\text{times}\}\\ =\frac{5}{9}\{\frac{10({10}^{\mathrm{n}}-1)}{10-1}-\mathrm{n}\}[âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}]\\ =\frac{5}{9}\{\frac{10({10}^{\mathrm{n}}-1)}{9}-\mathrm{n}\}\\ =\frac{50}{81}({10}^{\mathrm{n}}-1)-\frac{5\mathrm{n}}{9}\\ \left(\mathrm{ii}\right)\mathrm{Let}{\mathrm{S}}_{\mathrm{n}}=0.\text{6}+0.\text{66}+0.\text{666}+\dots \\ =\text{6}(0.1+0.11+111+1111+...)\\ =\frac{\text{6}}{9}(0.9+0.99+0.999+0.9999+...)\\ =\frac{\text{2}}{3}\{(1-0.1)+(1-0.01)+(1-0.001)+(1-0.0001)+...\}\\ =\frac{\text{2}}{3}\{1+1+1+1+...\mathrm{n}\text{times}-\frac{1}{10}-\frac{1}{{10}^2}-\frac{1}{{10}^3}-\frac{1}{{10}^4}-...\}\end{array}$ $\begin{array}{l}=\frac{\text{2}}{3}\{\mathrm{n}-\frac{\frac{1}{10}(1-\frac{1}{{10}^{\mathrm{n}}})}{1-\frac{1}{10}}\}[âˆ\mu {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}]\\ =\frac{\text{2}}{3}\mathrm{n}-\frac{\text{2}}{3}\times \frac{(1-\frac{1}{{10}^{\mathrm{n}}})}{9}\\ =\frac{\text{2}}{3}\mathrm{n}-\frac{\text{2}}{27}\times (1-\frac{1}{{10}^{\mathrm{n}}})\end{array}$

Q.80 Find the 20^th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Ans

$\begin{array}{l}\text{2}\times \text{4}+\text{4}\times \text{6}+\text{6}\times \text{8}+...+\text{n\hspace{0.17em}terms}\\ {\mathrm{T}}_{\mathrm{n}}=\{2+(\mathrm{n}-1)2\}\{4+(\mathrm{n}-1)2\}\\ =\left(2\mathrm{n}\right)(2\mathrm{n}+2)\\ =4\mathrm{n}(\mathrm{n}+1)\\ \mathrm{Putting}\text{n}=20,\text{we get}\\ {\mathrm{T}}_{20}=4\left(20\right)(20+1)\\ =80\times 21\\ =1680\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term of series is 1680.}\end{array}$

Q.81 Find the sum of the first n terms of the series: 3+ 7 + 13 + 21 + 31 +…

Ans

$\begin{array}{l}\mathrm{Let}{\text{S}}_{\text{n}}=3+7+13+21+31+...+{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}}\\ \mathrm{and}-\underset{¯}{{\text{S}}_{\text{n}}=-3-7-13-21-31-...-{\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}}}\\ 0=3+\left\{4+6+8+10+...\left(\mathrm{n}-1\right)\mathrm{terms}\right\}-{\mathrm{a}}_{\mathrm{n}}\\ {\mathrm{a}}_{\mathrm{n}}=3+\frac{\mathrm{n}-1}{2}\left\{8+\left(\mathrm{n}-2\right)2\right\}\\ =3+\left(\mathrm{n}-1\right)\left(4+\mathrm{n}-2\right)\\ =3+\left(\mathrm{n}-1\right)\left(\mathrm{n}+2\right)\\ {\mathrm{a}}_{\mathrm{n}}=3+{\mathrm{n}}^2+\mathrm{n}-2\\ ={\mathrm{n}}^2+\mathrm{n}+1\\ {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{n}}^2+\sum \mathrm{n}+\sum 1\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\\ =\frac{\mathrm{n}}{6}\left\{\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)+3\left(\mathrm{n}+1\right)+6\right\}\\ =\frac{\mathrm{n}}{6}\left(2{\mathrm{n}}^2+3\mathrm{n}+1+3\mathrm{n}+3+6\right)\\ =\frac{\mathrm{n}}{6}\left(2{\mathrm{n}}^2+6\mathrm{n}+10\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{3}\left({\mathrm{n}}^2+3\mathrm{n}+5\right)\end{array}$

Q.82

$\begin{array}{l}{\text{If S}}_{\text{1}}{\text{, S}}_{\text{2}}{\text{, S}}_{\text{3}}\text{are the sum of first n natural numbers,their squares and their cubes,respectively,}\\ \text{show that}9{\mathrm{S}}_2^2={\mathrm{S}}_3\left(1+8{\mathrm{S}}_1\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given that}\\ {\text{S}}_{\text{1}}=\sum \mathrm{n}\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\\ {\text{S}}_{\text{2}}=\sum {\mathrm{n}}^2\\ =\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\\ {\text{S}}_{\text{3}}=\sum {\mathrm{n}}^3\\ ={\left\{\frac{\mathrm{n}(\mathrm{n}+2)}{2}\right\}}^2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.={\mathrm{S}}_3(1+8{\mathrm{S}}_1)\\ ={\left\{\frac{\mathrm{n}(\mathrm{n}+2)}{2}\right\}}^2\{1+8\times \frac{\mathrm{n}(\mathrm{n}+1)}{2}\}\\ =\frac{{\mathrm{n}}^2{(\mathrm{n}+2)}^2}{4}\{1+4\mathrm{n}(\mathrm{n}+1)\}\\ =\frac{{\mathrm{n}}^2{(\mathrm{n}+2)}^2}{4}(1+4{\mathrm{n}}^2+4\mathrm{n})\\ =\frac{{\mathrm{n}}^2{(\mathrm{n}+2)}^2}{4}(4{\mathrm{n}}^2+4\mathrm{n}+1)\\ =\frac{{\mathrm{n}}^2{(\mathrm{n}+2)}^2{(2\mathrm{n}+1)}^2}{4}\\ ={\{3\times \frac{\mathrm{n}(\mathrm{n}+2)(2\mathrm{n}+1)}{6}\}}^2\end{array}$ $\begin{array}{l}=9{\left\{\frac{\mathrm{n}(\mathrm{n}+2)(2\mathrm{n}+1)}{6}\right\}}^2\\ =9{\mathrm{S}}_2^2=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.83

$\begin{array}{l}\mathbf{\text{Find the sum of the following series up to n terms:}}\\ \frac{{\mathbf{\text{1}}}^{\mathbf{\text{3}}}}{\mathbf{\text{1}}}\mathbf{\text{+}}\frac{{\mathbf{\text{1}}}^{\mathbf{\text{3}}}{\mathbf{\text{+2}}}^{\mathbf{\text{3}}}}{\mathbf{\text{1+3}}}\mathbf{\text{+}}\frac{{\mathbf{\text{1}}}^{\mathbf{\text{3}}}{\mathbf{\text{+2}}}^{\mathbf{\text{3}}}{\mathbf{\text{+3}}}^{\mathbf{\text{3}}}}{\mathbf{\text{1+3+5}}}\mathbf{\text{+}}\mathbf{.}\mathbf{..}\end{array}$

Ans

$\begin{array}{l}\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...\\ {\mathrm{T}}_{\mathrm{n}}=\frac{1^3+2^3+3^3+...+{\mathrm{n}}^3}{1+3+5+...}\\ =\frac{{\displaystyle \sum {\mathrm{n}}^3}}{\frac{\mathrm{n}}{2}\{2\times 1+(\mathrm{n}-1)2\}}\\ =\frac{{\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}}^2}{\frac{\mathrm{n}}{2}(2+2\mathrm{n}-2)}\\ =\frac{{\mathrm{n}}^2{(\mathrm{n}+1)}^2}{4{\mathrm{n}}^2}\\ =\frac{1}{4}({\mathrm{n}}^2+2\mathrm{n}+1)\end{array}$ $\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{4}(\sum {\mathrm{n}}^2+2\sum \mathrm{n}+\sum 1)\\ =\frac{1}{4}[\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\frac{2\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}]\\ =\frac{1}{4}\{\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\mathrm{n}+1+1\}\\ =\frac{\mathrm{n}}{4}\left(\frac{2{\mathrm{n}}^2+3\mathrm{n}+1+6\mathrm{n}+12}{6}\right)\\ =\frac{\mathrm{n}}{4}\left(\frac{2{\mathrm{n}}^2+9\mathrm{n}+13}{6}\right)\\ =\frac{\mathrm{n}}{24}(2{\mathrm{n}}^2+9\mathrm{n}+13)\end{array}$

Q.84

$\begin{array}{l}\mathrm{Show}\mathrm{that}\\ \frac{1\times 2^2+2\times 3^2+...+\mathrm{n}\times {(\mathrm{n}+1)}^2}{1^2\times 2+2^2\times 3+...+{\mathrm{n}}^2\times (\mathrm{n}+1)}=\frac{3\mathrm{n}+5}{3\mathrm{n}+1}\end{array}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1\times 2^2+2\times 3^2+...+\mathrm{n}\times {(\mathrm{n}+1)}^2}{1^2\times 2+2^2\times 3+...+{\mathrm{n}}^2\times (\mathrm{n}+1)}\\ =\frac{{\displaystyle \sum \{\mathrm{n}\times {(\mathrm{n}+1)}^2\}}}{{\displaystyle \sum \{{\mathrm{n}}^2\times (\mathrm{n}+1)\}}}\\ =\frac{{\displaystyle \sum ({\mathrm{n}}^3+2{\mathrm{n}}^2+\mathrm{n})}}{{\displaystyle \sum ({\mathrm{n}}^3+{\mathrm{n}}^2)}}\end{array}$ $\begin{array}{l}=\frac{{\displaystyle \sum {\mathrm{n}}^3+2\sum {\mathrm{n}}^2+\sum \mathrm{n}}}{{\displaystyle \sum {\mathrm{n}}^3+\sum {\mathrm{n}}^2}}\\ =\frac{{\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}}^2+2\left\{\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\right\}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{{\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}}^2+\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}}\\ =\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}+2\times \frac{2\mathrm{n}+1}{3}+1\}}{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{2\mathrm{n}+1}{3}\}}\\ =\frac{\left(\frac{3{\mathrm{n}}^2+3\mathrm{n}+8\mathrm{n}+4+6}{6}\right)}{\left(\frac{3{\mathrm{n}}^2+3\mathrm{n}+4\mathrm{n}+2}{6}\right)}\\ =\frac{3{\mathrm{n}}^2+11\mathrm{n}+10}{3{\mathrm{n}}^2+7\mathrm{n}+2}\\ =\frac{(\mathrm{n}+2)(3\mathrm{n}+5)}{(\mathrm{n}+2)(3\mathrm{n}+1)}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\frac{(3\mathrm{n}+5)}{(3\mathrm{n}+1)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.85 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Ans

$\begin{array}{l}\mathrm{Let}\text{150 workers finish a job in n days.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of workers started the job}=150\\ \text{Number of workers left the job per day}=4\\ \mathrm{The}\text{series of workers remain on the job every day}\\ 150,146,144,140,...\\ \mathrm{Then},\\ \mathrm{Number}\text{of workers finished job in a day}=\mathrm{Total}\mathrm{Number}\text{of}\\ \text{reducing workers finished job in}(\mathrm{n}+8)\mathrm{days}\\ \Rightarrow 150\mathrm{n}=150+146+144\\ +...(\mathrm{n}+8)\text{terms}\\ \Rightarrow 150\mathrm{n}=\frac{\mathrm{n}+8}{2}\left\{\begin{array}{l}2\times 150+\\ (\mathrm{n}+8-1)(-4)\end{array}\right\}\\ \Rightarrow 300\mathrm{n}=(\mathrm{n}+8)\{300-4(\mathrm{n}+7)\}\\ \Rightarrow 300\mathrm{n}=(\mathrm{n}+8)(300-4\mathrm{n}-28)\end{array}$ $\begin{array}{l}\Rightarrow 300\mathrm{n}=4(\mathrm{n}+8)(68-\mathrm{n})\\ \Rightarrow 75\mathrm{n}=-({\mathrm{n}}^2-60\mathrm{n}-544)\\ \Rightarrow {\mathrm{n}}^2-60\mathrm{n}-544+75\mathrm{n}=0\end{array}$ $\begin{array}{l}\Rightarrow {\mathrm{n}}^2+15\mathrm{n}-544=0\\ \Rightarrow (\mathrm{n}+32)(\mathrm{n}-17)=0\\ \Rightarrow \mathrm{n}=17,-32\left(\mathrm{Negative}\right)\\ \mathrm{Thus},\text{the number of days required to complete work by 150}\\ \text{workers is 17 days.Therefore, number of days required to finish}\\ \text{the job by reducing number of workers}=18+7\\ =25\mathrm{days}\end{array}$