Important Questions Class 11 Maths Chapter 9

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 Question-answer format. These concise questions are prepared by subject matter experts by referring to NCERT books.

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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