 # Important Questions for Class 11 Maths Chapter 9 - Sequences and Series

Important questions for class 11 Maths Chapter 9 – sequences and series are given here. Sequences and series have several applications in our daily life. The important questions cover all the topics and subtopics in the NCERT textbook. The solutions for the questions are given in a step by step procedure so that students can understand them in a better way. Go through the important questions provided at BYJU’S and achieve excellent marks in the annual examination. Get all the important Maths questions from class 11 chapters here.

Class 11 Maths Chapter 9 – Sequences and series cover the following important concepts such as:

• Introduction to Sequences and Series
• Arithmetic Progression
• Geometric Progression
• Sum of n terms of the special series
• The relation between the arithmetic and geometric mean

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## Class 11 Chapter 9 – Sequences and Series Important Questions with Solutions

Practice class 11 chapter 9 sequences and series problems provided here, which are taken from the previous year question papers.

Question 1:

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

Solution:

Let a1, a2 and d1, d2 be the first term and the common difference of the first and second

arithmetic progression respectively.

Then,

(Sum of n terms of the first A.P)/(Sum of n terms of the second A.P) = (5n+4)/(9n+6)

⇒ [ (n/2)[2a1+ (n-1)d1]]/ [(n/2)[2a2+ (n-1)d2]]= (5n+4)/(9n+6)

Cancel out (n/2) both numerator and denominator on L.H.S

⇒[2a1+ (n-1)d1]/[2a2+ (n-1)d2]= (5n+4)/(9n+6) …(1)

Now susbtitute n= 35 in equation (1), {Since (n-1)/2 = 17}

Then equation (1) becomes

⇒[2a1+ 34d1]/[2a2+ 34d2]= (5(35)+4)/(9(35+6)

⇒[a1+ 17d1]/[a2+ 17d2]= 179/321 …(2)

Now, we can say that.

18th term of first AP/ 18th term of second AP = [a1+ 17d1]/[a2+ 17d2]….(3)

Now, from (2) and (3), we can say that,

18th term of first AP/ 18th term of second AP = 179/321

Hence, the ratio of the 18th terms of both the AP’s is 179:321.

Question 2:

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

Solution:

Assume that A1, A2, A3, A4, and A5 are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A1, A2, A3, A4, A5, 26

Here, a= 8, l =26, n= 5

Therefore, 26= 8+(7-1)d

Hence it becomes,

26 = 8+6d

6d = 26-8 = 18

6d= 18

d = 3

A1= a+d = 8+ 3 =11

A2= a+2d = 8+ 2(3) =8+6 = 14

A3= a+3d = 8+ 3(3) =8+9 = 17

A4= a+4d = 8+ 4(3) =8+12 = 20

A5= a+5d = 8+ 5(3) =8+15 = 23

Hence, the required five numbers between the number 8 and 26 are 11, 14, 17, 20, 23.

Question 3:

The 5th, 8th, and 11th terms of a GP are p, q and s respectively. Prove that q2 = ps

Solution:

Given that:

5th term = P

8th term = q

11th term = s

To prove that: q2 = ps

By using the above information, we can write the equation as:

a5 = ar5-1 = ar4 = p ….(1)

a8 = ar8-1 = ar7 = q ….(2)

a11 = ar11-1 = ar10 =s …(3)

Divide the equation (2) by (1), we get

r3 = q/p …(4)

Divide the equation (3) by (2), we get

r3 = s/q …(5)

Now, equate the equation (4) and (5), we get

q/p = s/q

It becomes, q2 = ps

Hence proved.

Question 4:

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

Solution:

Let a and d be the first term and the common difference of the A.P. respectively. It is known

that the kth term of an A.P. is given by

ak = a +(k -1)d

Therefore, am+n = a +(m+n -1)d

am-n = a +(m-n -1)d

am = a +(m-1)d

Hence, the sum of (m + n)th and (m – n)th terms of an A.P is written as:

am+n+ am-n = a +(m+n -1)d + a +(m-n -1)d

= 2a +(m + n -1+ m – n -1)d

=2a+(2m-2)d

=2a + 2(m-1)d

= 2 [a + (m-1)d]

= 2 am [since am = a +(m-1)d]

Therefore, the sum of (m + n)thand (m – n)th terms of an A.P. is equal to twice the mth term.

Question 5:

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

Solution:

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 ….. 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒ 100=2+(n-1)2

⇒ n= 50

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]

= (50/2)(4+98)

= 25(102)

= 2550

The integers from 1 to 100, which are divisible by 5, 10…. 100

This forms an A.P. with both the first term and common difference equal to 5.

Therefore, 100= 5+(n-1)5

⇒5n = 100

⇒ n= 100/5

⇒ n= 20

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]

= (20/2)(10+95)

= 10(105)

= 1050

Hence, the integers from 1 to 100, which are divisible by both 2 and 5 are 10, 20, ….. 100.

This also forms an A.P. with both the first term and common difference equal to 10.

Therefore, 100= 10+(n-1)10

⇒10n = 100

⇒ n= 100/10

⇒ n= 10

10+20+…+100= (10/2)[2(10)+(10-1)(10)]

= (10/2)(20+90)

= 5(110)

= 550

Therefore, the required sum is:

= 2550+ 1050 – 550

= 3050

Hence, the sum of the integers from 1 to 100, which are divisible by 2 or 5 is 3050.

### Practice Problems for Class 11 Maths Chapter 9 Sequences and Series

Go through class 11 chapter 9 sequences and series concepts and solve the practice problems provided below:

1. Prove that the sequence tn = 3n + 5 is an Arithmetic Progression. Find its common difference.
2. Find the 10th term which is common between the given AP’s.
1. 3+7+ 11+…..
2. 1 + 6 + 11 +16
3. Is 309 a term of the AP 11, 17, 23….?
4. The sum of 3 numbers of an arithmetic progression is 24 and their product is 44. Find the three numbers.
5. If the sum of first n terms of a progression is a quadratic expression in n, show that it is an arithmetic progression.
6. If the AM between the ath and bth terms of an AP be equal to the AM between the cth and dth term of it then prove that a + b = c + d
7. Find three numbers in GP whose product is 1728 and sum is 38.
8. What will Rs 5000 amount to in 10 years, compounded annually at 10 % per annum?
9. Find two positive numbers m and n whose AM and GM are 34 and 16 respectively.
10. Find the sum of the series – 1 + 5 + 12 +….to n terms.
11. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year, he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?
12. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
13. Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + … to
(i) n terms
(ii) 10 terms

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