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Important Questions Class 11 Mathematics Chapter 9
Important Questions for CBSE Class 11 Mathematics Chapter 9 – Sequences and Series
Class 11 Mathematics Chapter 9 Sequences and Series explains sequences, arithmetic progression, geometric progression, the Fibonacci series, and the sum of specific natural number sequences involving squares and cube roots. A logical explanation has also been provided for the problem of arithmetic and geometric mean.
Apart from the introductory section in which the entire chapter is introduced, Chapter 9 of Class 11 Mathematics is divided into six important sections. It goes on to explain the concepts of Sequence, Series, Arithmetic, and Geometric Progression, as well as the correlation between them and the sum related to n terms of special series.
The Sequence and Series chapter requires a lot of practice to strengthen conceptual understanding because it contains many difficult sections. The underlying concepts can be challenging at first, but with the right kind of guidance and advanced techniques, students can score the highest possible marks in their exams.
Extramarks Important Questions for Class 11 Mathematics Chapter 9 is useful for students who want to study this chapter in a Questionanswer format. These concise questions are prepared by subject matter experts by referring to NCERT books.
CBSE Class 11 Mathematics Chapter9 Important Questions – Free Download
(add important questions)
Study Important Questions for Class 12 Mathematics Chapter 9 – Sequences and Series
Given below are some of the Chapter 9 Class 11 Mathematics Important Questions. Students can click on the link given to access the complete set of questions.
Short Answer Questions – 2 Marks
Q1. The third term of a geometric progression is 4. When the first five terms are multiplied together, what is the result?
Ans: T 3 = 4 is deduced from the given question.
⇒ ar2 = 4
The product of the first five terms can now be defined as = a.ar.ar2.ar3.ar4.
= a5r10
= (ar2)5
= 4^5
As a result, the product of the first five terms is 45.
Q2. 13/6 is the sum of two numbers. An even number of mathematical means are placed between them, and their sum exceeds their number by one. What is the number of inserted means multiplied by two?
Assume a and b are two numbers such that
a + b = 13/6
Take A1, A2, A3,………..
A2n is the arithmetic mean of a and b.
Then, A1 + A2 + A3 +………..+ A2n = 2n(n + 1)/2
⇒ n(a + b) = 13n/6
Given this, the series A1 + A2 + A3 +………..+ A2n = 2n + 1
13n/6 = 2n + 1, so
⇒ n = 6
Short Answer Questions – 4 Marks
Q1. A total of 150 workers were hired to finish a job in a specific number of days. Four workers dropped out on the second day, four more dropped out on the third day, and so on. Determine the number of days it took to complete the task and the number of days it took to complete the work.
Ans: A = 150, d = 4
Sn= n2[2×150+(n−1)(−4)]
If the total number of workers who worked for the entire n days was 150, (n – 8)
∴n2[300 + (n – 1)( – 4)] = 150(n – 8) (n – 8)
⇒n = 25
Prove that the sum of n terms of the series 11 + 10^3 + 100^5 + …..is10/9(10^n – 1) + n^2.
Ans: Here, using the formula,
Sn = 11 + 10^3 + 100^5 + …… + n terms
Sn = (10 + 1) + (10^2 + 3) + (10^3 + 5) + …. + (10^n+(2^n – 1))
Sn = 10(10^n – 1)10 – 1 + n^2(1 + 2^n – 1)
=10/9(10^n – 1) + n^2
Long Answer Questions – 6 Marks
Q1. If the sum of two numbers is 6 times their geometric mean, show that the numbers are in the ratio 3+32–√):(3−22–√) .
A1. a + b = 6ab−−√
a + b2ab−−√=31
By C and D, we get,
a + b + 2ab−−√a + b – 2ab−−√=3+13−1
⇒(a–√ + b−−√)2(a–√ – b−−√)2 = 21
⇒a–√ + b−−√a–√ – b−−√ = 2–√1
Again by C and D, we get,
a–√ + b−−√ + a–√ – b−−√a–√ + b−−√−a–√+b−−√=2–√+12–√−1
⇒2a–√2b−−√=2–√+12–√−1
⇒a/b = (2–√ + 1)2(2–√ – 1)2 (Squaring Both Sides)
⇒a/b = 2+1+22–√2+1−22–√
⇒a/b = 3+22–√3−22–√
⇒a:b = (3 + 22–√):(3 – 22–√)
Q2. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. It has a 5:9 ratio of 7th to (m – 1)th numbers. Determine the value of m.
A2. Let 1, A1, A2, …., Am, 31 be in A.P.
a = 1, an = 31
am+2 = 314
an = a + (n – 1)d
31 = a + (m + 2 – 1)d
d = 30m + 1
A7Am – 1 = 59(given)
⇒1 + 7(30m + 1)1 + (m – 1)(30m + 1) = 59
⇒m = 1
Q3. 150 workers were hired to complete a job in a certain number of days. Four workers left on the second day, four more on the third day, and so on. It took 8 more days to complete the work; what was the number of days it took to complete the work?
A3. A = 150, d = 4
Sn= n2[2×150+(n−1)(−4)]
If the total workers who would have worked for all n days, 150(n – 8)
∴n2[300 + (n – 1)( – 4)] = 150(n – 8)
⇒n = 25
Q4. Show that the sum of the first n terms of the series 11 + 103 + 1005 + ….. equals 109 (10n – 1) + n2.
A4. Sn = 11 + 103 + 1005 + …… + n terms
Sn = (10 + 1) + (102 + 3) + (103 + 5) + …. + (10n+(2n – 1))
Sn = 10(10n – 1)10 – 1 + n2(1 + 2n – 1)
=109(10n – 1) + n2
Q.1 The sum of first 24 terms of the A.P. a_{1}, a_{2}, a_{3},, if it is given that a_{1}+ a_{5}+ a_{10}+ a_{15}+ a_{20}+ a_{24 }= 225, is
Marks:1
Ans 900.
We know that in an A.P. the sum of the terms equidistance from the beginning and end is always same and is equal to the sum of first and last term i.e.,
a_{1}+ a_{n }= a_{2}+ a_{n1}= a_{3}+ a_{n2 }=
So, if an A.P. consists of 24 terms, then
a_{1}+ a_{24}= a_{5}+ a_{20}= a_{10}+ a_{15}.
Now, a_{1}+ a_{5}+ a_{10}+ a_{15}+ a_{20}+ a_{24 }= 225
i.e.,(a_{1}+ a_{24})+(a_{5}+a_{15})+( a_{10}+ a_{15}) = 225
3(a_{1}+ a_{24}) = 225
a_{1}+ a_{24} = 75 (1)
Therefore, S_{24 }= (24/2) (a_{1}+ a_{24})
= 900.
Q.2 The fourth term of a G.P. is 6, then the product of its first eight terms is
Marks:1
Ans
Let a be the first term and r the common ratio.
Therefore, ar^{3}= 6 (1)
Now, product of first eight terms
a(ar)(ar^{2})(ar^{3})(ar^{4}) (ar^{5}) (ar^{6}) = a^{7}r^{21}
= (ar^{3})^{7}
= 6^{7}.
Q.3 Find the sum of the series:0.7 + 0.77 + 0.777 + to n terms
Marks:6
Ans
We have 0.7+0.77+0.777+ to n terms=710+77100+7771000+ to n terms=7110+11100+1111000+ to n terms On multiply and divide by 9 we get, =79910+99100+9991000+ to n terms=7910110+1001100+100011000+ to n terms=791110+11100+111000+ to n terms=791+1+1+ to n terms110+1100+11000+ to n terms=79n110+1102+1103+ to n terms Here, we get GP with a=110 and r=110Sn=a1rn1r=79n1101110n1110=79n110—1091110n1=79n191110n
Q.4 If a, b, c are in G.P., then prove that a^{2} b^{2}, b^{2} c^{2}, c^{2} a^{2} are in G.P.
Marks:4
Ans
a, b, c are in G.P. Let r be the common ratio. Then
b = ar and c = ar^{2}
a^{2} b^{2}, b^{2} c^{2}, c^{2} a^{2} will be in G.P. if
b2c2a2b2=c2a2b2c2
Substituting b = ar and c = ar^{2} we get
ar2ar22a2ar2 =ar22ar32ar2ar22a2r21r2a21r2 =a2r41r2a2r21r2r2 =r2
Hence a^{2} b^{2}, b^{2} c^{2}, c^{2} a^{2} is a G.P.
Q.5 A snail travels in a straight line path. Assume that it travels 1cm in the first second and half of the distance traveled in the previous second in the succeeding second. In how many seconds would it reach a point 31/16 cm away Thus calculate the speed of the snail.
Marks:6
Ans
Distance traveled by snail in the first second = 1 cm
Distance traveled in the second sec.
=1—12=12cm
Distance traveled in the third sec
=12—12=14cm
Thus the distances covered form a G.P. 1, 1/2, 1/4,…
Here, a = 1 and r
=12.
Let n seconds be the required time. Therefore,
Sum =1112n112=311612n=132n=5secnow, speed=distance· time=3116·5=3180cm/sec
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FAQs (Frequently Asked Questions)
1. What are sequence and series?
In mathematics, a sequence is defined as a list of objects ordered sequentially, such that each member comes before or after every other member. A series is defined as the sum of terms in a sequence.
2. What is the difference between sequence series and progression?
The main distinction between sequence and progression is that a sequence is based on a logical rule and is not linked to a formula. Whereas progression is determined by a formula.
3. What does Arithmetic Progression consist of?
Progression is a sequence in which the terms follow a specific pattern. In the case of Arithmetic progression, there are two consecutive terms with a constant difference.
4. What are the 4 Types of Sequences?
The 4 types of sequences are:
 Arithmetic Sequences
 Geometric Sequences
 Harmonic Sequences
 Fibonacci Numbers
5. What are some examples of common sequence types?
The following are some examples of common sequences:
Arithmetic Progression (A.P) is a sequence in which the terms increase and decrease by the same constant, known as the A.P’s common difference. The first term of an A.P is usually denoted by ‘a,’ the common difference by ‘d,’ and the final term by ‘l’.
Geometric Progression (G.P) is a sequence in which the ratio of each term to the term before it remains constant throughout. The first term of the G.P is denoted by ‘a’, and the common ratio is denoted by ‘r’.