NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.2

Q.1 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.2 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.3 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.

$\begin{array}{l}\mathrm{First}\text{term of A.P.}=\text{2}\\ \text{Let common difference}=\mathrm{d}\\ \mathrm{According}\text{to given condition:}\\ \text{Sum of first 5 terms}=\frac{1}{4}\left(\mathrm{Sum}\text{of next 5 terms}\right)\\ {\mathrm{S}}_5=\frac{1}{4}({\mathrm{S}}_{10}-{\mathrm{S}}_5)\\ \Rightarrow 4{\mathrm{S}}_5={\mathrm{S}}_{10}-{\mathrm{S}}_5\\ \Rightarrow 5{\mathrm{S}}_5={\mathrm{S}}_{10}\end{array}$

$\begin{array}{l}\Rightarrow 5\left[\frac{5}{2}\{2\times 2+(5-1)\mathrm{d}\}\right]=\left[\frac{10}{2}\{2\times 2+(10-1)\mathrm{d}\}\right]\\ [\because {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}]\\ \Rightarrow \frac{25}{2}(4+4\mathrm{d})=5(4+9\mathrm{d})\\ \Rightarrow 25(2+2\mathrm{d})=5(4+9\mathrm{d})\\ \Rightarrow 50+50\mathrm{d}=20+45\mathrm{d}\\ \Rightarrow 5\mathrm{d}=20-50\\ \Rightarrow \mathrm{d}=-\frac{30}{5}\\ =-6\\ \mathrm{Now},\text{sum of 20 terms.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{20}}=2+(20-1)\times -6[\because {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}]\\ =2-114\\ =-112\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term is}-\text{112.}\end{array}$

Q.4

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

Ans.

$\begin{array}{l}\mathrm{The}\text{given A.P. is}-6,-\frac{11}{2},-5,...\\ \mathrm{Let}\text{the sum of n terms of A.P. is}-\text{25.Then,}\end{array}$

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

Q.5

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

Ans.

$\begin{array}{l}\mathrm{Let}\text{first term of A.P.}=\mathrm{a}\\ \mathrm{Let}\text{common difference of A.P.}=\mathrm{d}\\ \mathrm{Since},{\text{ T}}_{\text{p}}=\frac{1}{\mathrm{q}}\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{a}+(\mathrm{p}-1)\mathrm{d}=\frac{1}{\mathrm{q}}...\left(\mathrm{i}\right)\\ \mathrm{And},{\text{ T}}_{\text{q}}=\frac{1}{\mathrm{p}}\\ \Rightarrow \mathrm{a}+(\mathrm{q}-1)\mathrm{d}=\frac{1}{\mathrm{p}}...\left(\mathrm{ii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ (\mathrm{p}-1-\mathrm{q}+1)\mathrm{d}=\frac{1}{\mathrm{q}}-\frac{1}{\mathrm{p}}\\ \Rightarrow (\mathrm{p}-\mathrm{q})\mathrm{d}=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{pq}}\\ \Rightarrow \mathrm{d}=\frac{1}{\mathrm{pq}}\\ \mathrm{Substituting}\text{value of d in equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{a}+(\mathrm{p}-1)\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ \Rightarrow \mathrm{a}+\frac{\mathrm{p}}{\mathrm{pq}}-\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ \Rightarrow \mathrm{a}=\frac{1}{\mathrm{pq}}\\ \mathrm{Now},{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\{2\times \frac{1}{\mathrm{pq}}+(\mathrm{pq}-1)\frac{1}{\mathrm{pq}}\}\\ [\because {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\}]\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\end{array}$

$\begin{array}{l}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\{2\times \frac{1}{\mathrm{pq}}+(\mathrm{pq}-1)\frac{1}{\mathrm{pq}}\}\\ =\frac{\mathrm{pq}}{2}(\frac{2}{\mathrm{pq}}+1-\frac{1}{\mathrm{pq}})\\ =\frac{\mathrm{pq}}{2}(\frac{1}{\mathrm{pq}}+1)\\ =\frac{1}{2}(1+\mathrm{pq})\\ \mathrm{Thus},\text{the sum of pq terms is}\frac{1}{2}(\mathrm{pq}+1).\end{array}$

Q.6 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{\hspace{0.33em}S}}_{\text{n}}=116\\ \frac{\text{n}}{2}\left\{2\times 25+\left(\text{n}-1\right)\left(-3\right)\right\}=116\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}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{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n}=8,\frac{29}{3}\end{array}\\ \mathrm{Since},\text{n is a natural number.So,}\\ \begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}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.7 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.8 If the sum of n terms of an A.P. is (pn + qn²), where p and q are constants, find the common difference.

Ans.

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

$\begin{array}{l}=\mathrm{p}+3\mathrm{q}\\ {\mathrm{T}}_2={\mathrm{S}}_3-{\mathrm{S}}_2\\ =(3\mathrm{p}+9\mathrm{q})-(2\mathrm{p}+4\mathrm{q})\\ =\mathrm{p}+5\mathrm{q}\\ \mathrm{Common}\text{ difference,}\\ \text{d}={\mathrm{T}}_2-{\mathrm{T}}_1\\ =(\mathrm{p}+5\mathrm{q})-(\mathrm{p}+3\mathrm{q})\\ =2\mathrm{q}\\ \mathrm{Thus},\text{common difference is 2q.}\end{array}$

Q.9 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.10 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.11

$\begin{array}{l}\text{Sum of the first p, q and r terms of an A.P. are a, b and c, \hspace{0.33em}respectively.}\\ \text{Prove that}\frac{\text{a}}{\text{p}}\left(\text{q}-\text{r}\right)\text{+}\frac{\text{b}}{\text{q}}\left(\text{r}-\text{p}\right)\text{+}\frac{\text{c}}{\text{r}}\left(\text{p}-\text{q}\right)\text{\hspace{0.33em}=\hspace{0.33em}0.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be A and common difference be d.}\\ \mathrm{Given}:{\mathrm{S}}_{\mathrm{p}}=\mathrm{a}\Rightarrow \mathrm{a}=\frac{\mathrm{p}}{2}\{2\mathrm{A}+(\mathrm{p}-1)\mathrm{d}\}\\ \Rightarrow \frac{\mathrm{a}}{\mathrm{p}}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{p}-1)\mathrm{d}\}\\ {\mathrm{S}}_{\mathrm{q}}=\mathrm{b}\Rightarrow \mathrm{b}=\frac{\mathrm{p}}{2}\{2\mathrm{A}+(\mathrm{q}-1)\mathrm{d}\}\\ \Rightarrow \frac{\mathrm{b}}{\mathrm{q}}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{q}-1)\mathrm{d}\}\\ \text{and}{\mathrm{S}}_{\mathrm{r}}=\mathrm{c}\Rightarrow \mathrm{c}=\frac{\mathrm{r}}{2}\{2\mathrm{A}+(\mathrm{r}-1)\mathrm{d}\}\\ \Rightarrow \frac{\mathrm{c}}{\mathrm{r}}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{r}-1)\mathrm{d}\}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}}{\mathrm{p}}(\mathrm{q}-\mathrm{r})+\frac{\mathrm{b}}{\mathrm{q}}(\mathrm{r}-\mathrm{p})+\frac{\mathrm{c}}{\mathrm{r}}(\mathrm{p}-\mathrm{q})\\ \end{array}$

$\begin{array}{l}=\frac{1}{2}\{2\mathrm{A}+(\mathrm{p}-1)\mathrm{d}\}(\mathrm{q}-\mathrm{r})+\frac{1}{2}\{2\mathrm{A}+(\mathrm{q}-1)\mathrm{d}\}(\mathrm{r}-\mathrm{p})\\ +\frac{1}{2}\{2\mathrm{A}+(\mathrm{r}-1)\mathrm{d}\}(\mathrm{p}-\mathrm{q})\\ =\frac{1}{2}\left[\begin{array}{l}2\mathrm{A}(\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q})+\\ \mathrm{d}\{(\mathrm{p}-1)(\mathrm{q}-\mathrm{r})+(\mathrm{q}-1)(\mathrm{r}-\mathrm{p})+(\mathrm{r}-1)(\mathrm{p}-\mathrm{q})\}\end{array}\right]\\ =\frac{1}{2}\{2\mathrm{A}\left(0\right)+\mathrm{d}\left(\begin{array}{l}\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{qp}-\mathrm{r}+\mathrm{p}\\ +\mathrm{rp}-\mathrm{rq}-\mathrm{p}+\mathrm{q}\end{array}\right)\}\\ =\frac{1}{2}(0+0)\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.12 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.13 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.

$\begin{array}{l}\mathrm{It}\text{\hspace{0.33em}is given that}\\ {\text{S}}_{\text{n}}={\text{3n}}^{\text{2}}+\text{5n}\\ \text{Replace n by n}-\text{1, we get}\\ {\text{S}}_{\text{n-1}}=\text{3}{\left(\text{n}-\text{1}\right)}^{\text{2}}+\text{5}\left(\text{n}-1\right)\\ ={\text{3n}}^{\text{2}}-6\text{n}+\text{3}+\text{5n}-5\\ ={\text{3n}}^{\text{2}}-\text{n}-2\\ \mathrm{So},{\text{\hspace{0.33em}T}}_{\text{n}}={\text{S}}_{\text{n}}-{\text{S}}_{\text{n-1}}\\ =\left({\text{3n}}^{\text{2}}+\text{5n}\right)-\left({\text{3n}}^{\text{2}}-\text{n}-2\right)\\ =6\text{n}+2\\ \mathrm{Since},{\text{\hspace{0.33em}\hspace{0.33em}T}}_{\text{m}}=164\\ \mathrm{So},\text{\hspace{0.33em}6m}+\text{2}=\text{164}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}m}=\frac{164-2}{6}\\ =\frac{162}{6}\\ =27\\ \mathrm{Thus},\text{the value of m is 27.}\end{array}$

Q.14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Ans.

$\begin{array}{l}\mathrm{Let}{\text{5 numbers between 8 and 26 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{and A}}_{\text{5}}\text{.}\\ {\text{Then, A.P. is 8, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{, A}}_{\text{5}}\text{,\hspace{0.17em}26.}\\ \because \mathrm{Last}\text{term}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 26=8+(7-1)\mathrm{d}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}26=8+6\mathrm{d}\\ \mathrm{d}=\frac{26-8}{6}\\ =\frac{18}{6}\\ =3\\ \mathrm{Then},{\mathrm{A}}_1=8+\mathrm{d}\\ =8+3\\ =11\\ {\mathrm{A}}_2=8+2\mathrm{d}\\ =8+2\left(3\right)\\ =14\\ {\mathrm{A}}_3=8+3\mathrm{d}\\ =8+3\left(3\right)\\ =17\\ {\mathrm{A}}_4=8+4\mathrm{d}\\ =8+4\left(3\right)\\ =20\end{array}$ \begin{array}{l}\mathrm{}\end{array} $\begin{array}{l}\end{array}$

$\begin{array}{l}{\mathrm{A}}_5=8+5\mathrm{d}\\ =8+5\left(3\right)\\ =23\\ \mathrm{Thus},\text{the 5 terms between 8 and 26 are 11, 14, 17, 20 and 23.}\end{array}$

Q.15

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.16 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.

$\begin{array}{l}\mathrm{Let}{\text{m numbers between 1 and 31 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{.}\\ {\text{Then, A.P. is 1, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{,\hspace{0.17em}\hspace{0.17em}31.}\\ \because \mathrm{Last}\text{term}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ 31=1+(\mathrm{m}+2-1)\mathrm{d}\\ 30=(\mathrm{m}+1)\mathrm{d}\\ \mathrm{d}=\frac{30}{\mathrm{m}+1}\\ \mathrm{Then},{\mathrm{A}}_1=1+\mathrm{d}\\ =1+\frac{30}{\mathrm{m}+1}\\ =\frac{\mathrm{m}+31}{\mathrm{m}+1}\\ {\mathrm{A}}_7=1+7\mathrm{d}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}=1+7\left(\frac{30}{\mathrm{m}+1}\right)\\ =\frac{\mathrm{m}+211}{\mathrm{m}+1}\\ {\mathrm{A}}_{\mathrm{m}-1}=1+(\mathrm{m}-1)\mathrm{d}\\ =1+(\mathrm{m}-1)\left(\frac{30}{\mathrm{m}+1}\right)\\ =\frac{\mathrm{m}+1+30(\mathrm{m}-1)}{\mathrm{m}+1}\\ =\frac{31\mathrm{m}-29}{\mathrm{m}+1}\\ \mathrm{According}\text{to question:}\\ \frac{{\mathrm{A}}_7}{{\mathrm{A}}_{\mathrm{m}-1}}=\frac{\left(\frac{\mathrm{m}+211}{\mathrm{m}+1}\right)}{\left(\frac{31\mathrm{m}-29}{\mathrm{m}+1}\right)}\\ \frac{5}{9}=\frac{\mathrm{m}+211}{31\mathrm{m}-29}\\ 155\mathrm{m}-145=9\mathrm{m}+1899\\ 155\mathrm{m}-9\mathrm{m}=1899+145\\ 146\mathrm{m}=2044\\ \mathrm{m}=\frac{2044}{146}\\ =14\\ \mathrm{Thus},\text{the value of m is 14.}\end{array}$

Q.17 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.18

Ans.

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