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

Chapter 9 Sequences and Series of Class 11 Maths is categorized under the term â€“ I CBSE Syllabus for 2021-22. The second exercise of this chapter is based on the topic of Arithmetic Progression and arithmetic mean. Arithmetic Progression (A.P.) or arithmetic sequence is a sequence of numbers such that the difference between consecutive terms is constant. The exercise also deals with questions related to Arithmetic Mean, which is defined as, the average of a set of numerical values, as calculated by adding them together and dividing by the number of terms in the set. Learn and understand more about these topics by solving Exercise 9.2 of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series.

There are a lot of questions given in the NCERT textbook for the students to solve and practice. Solving the NCERT Solutions for Class 11 Maths and practising will undoubtedly aid the students in scoring high marks in the Class 11 term â€“ I examinations. Students have to ensure that they practise every problem given in the textbook repeatedly until the concept gets clear.

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### Access Other Exercise Solutions for NCERT Class 11 Maths Chapter 9- Sequences and Series

Exercise 9.1 Solutions 14 Questions

Exercise 9.3 Solutions 32 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise On Chapter 9 Solutions 32 Questions

#### Access Solutions for NCERT Class 11 Maths Chapter 9 Exercise 9.2

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

Solution:

The odd integers from 1 to 2001 are 1, 3, 5, â€¦1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term,Â aÂ = 1

Common difference,Â dÂ = 2

Now,

a + (n -1)d = 2001

1 + (n-1)(2) = 2001

2n â€“ 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of odd numbers from 1 to 2001 is 1002001.

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

Solution:

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, â€¦ 995.

It clearly forms a sequence in A.P.

Where, the first term,Â aÂ = 105

Common difference,Â dÂ = 5

Now,

a + (n -1)d = 995

105 + (n â€“ 1)(5) = 995

105 + 5n â€“ 5 = 995

5n = 995 â€“ 105 + 5 = 895

n = 895/5

n = 179

We know,

Sn = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

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 20thÂ term is â€“112.

Solution:

Given,

The first term (a) of an A.P = 2

Letâ€™s assumeÂ dÂ be the common difference of the A.P.

So, the A.P. will be 2, 2 +Â d, 2 + 2d, 2 + 3d, â€¦

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = Â¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20 â€“ 1)d = 2 + (19) (-6) = 2 â€“ 114 = -112

Therefore, the 20thÂ term of the A.P. is â€“112.

4. How many terms of the A.P. -6, -11/2, -5, â€¦.Â are needed to give the sum â€“25?

Solution:

Letâ€™s consider the sum ofÂ nÂ terms of the given A.P. as â€“25.

We known that,

Sn = n/2 [2a + (n-1)d]

whereÂ nÂ = number of terms,Â aÂ = first term, andÂ dÂ = common difference

So here,Â aÂ = â€“6

d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2

Thus, we have

5. In an A.P., ifÂ pthÂ term isÂ 1/q andÂ qthÂ term isÂ 1/p, prove that the sum of firstÂ pqÂ terms is Â½ (pq + 1) where p â‰  q. Â

Solution:

6. If the sum of a certain number of terms of the A.P. 25, 22, 19, â€¦ is 116. Find the last term

Solution:

Given A.P.,

25, 22, 19, â€¦

Here,

First term, a = 25 and

Common difference, d = 22 â€“ 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

Sn = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

116 x 2 = n [50 â€“ 3n + 3]

232 = n [53 â€“ 3n]

232 = 53n â€“ 3n2

3n2 â€“ 53n + 232 = 0

3n2 â€“ 24n â€“ 29n+ 232 = 0

3n(n â€“ 8) â€“ 29(n â€“ 8) = 0

(3n â€“ 29) (n â€“ 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, 8th term is the last term of the A.P.

a8 = 25 + (8 â€“ 1)(-3)

= 25 â€“ 21

= 4

7. Find the sum toÂ nÂ terms of the A.P., whoseÂ kthÂ term is 5kÂ + 1.

Solution:

Given, theÂ kthÂ term of the A.P. is 5kÂ + 1.

kthÂ term =Â akÂ =Â aÂ + (kÂ â€“ 1)d

And,

aÂ + (kÂ â€“ 1)dÂ = 5kÂ + 1

aÂ +Â kdÂ â€“Â dÂ = 5kÂ + 1

On comparing the coefficient ofÂ k, we getÂ dÂ = 5

aÂ â€“Â dÂ = 1

aÂ â€“ 5 = 1

â‡’Â aÂ = 6

8. If the sum ofÂ nÂ terms of an A.P. is (pnÂ +Â qn2), whereÂ pÂ andÂ qÂ are constants, find the common difference.

Solution:

We know that,

Sn = n/2 [2a + (n-1)d]

From the question we have,

On comparing the coefficients ofÂ n2Â on both sides, we get

d/2 = q

Hence, dÂ = 2q

Therefore, the common difference of the A.P. is 2q.

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

Solution:

LetÂ a1,Â a2, andÂ d1,Â d2Â be the first terms and the common difference of the first and second arithmetic progression respectively.

Then, from the question we have

10. If the sum of firstÂ pÂ terms of an A.P. is equal to the sum of the firstÂ qÂ terms, then find the sum of the first (pÂ +Â q) terms.

Solution:

Letâ€™s takeÂ aÂ andÂ dÂ to be the first term and the common difference of the A.P. respectively.

Then, it given that

Therefore, the sum of (p + q) terms of the A.P. is 0.

11. Sum of the firstÂ p, qÂ andÂ rÂ terms of an A.P. areÂ a, bÂ andÂ c, respectively.

Prove thatÂ

Solution:

LetÂ a1Â andÂ dÂ be the first term and the common difference of the A.P. respectively.

Then according to the question, we have

Now, subtracting (2) from (1), we get

12. The ratio of the sums ofÂ mÂ andÂ nÂ terms of an A.P. isÂ m2:Â n2. Show that the ratio ofÂ mthÂ andÂ nthÂ term is (2mÂ â€“ 1): (2nÂ â€“ 1).

Solution:

Letâ€™s consider thatÂ aÂ andÂ bÂ to be the first term and the common difference of the A.P. respectively.

Then from the question, we have

Hence, the given result is proved.

13. If the sum ofÂ nÂ terms of an A.P. isÂ 3n2 + 5n and itsÂ mthÂ term is 164, find the value ofÂ m.

Solution:

Letâ€™s considerÂ aÂ andÂ bÂ to be the first term and the common difference of the A.P. respectively.

amÂ =Â aÂ + (mÂ â€“ 1)dÂ = 164 â€¦ (1)

We the sum of the terms is given by,

Sn = n/2 [2a + (n-1)d]

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

Solution:

Letâ€™s assume A1, A2, A3, A4, and A5Â to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.

Here we have,

aÂ = 8,Â bÂ = 26,Â nÂ = 7

So,

26 = 8 + (7 â€“ 1)Â d

6dÂ = 26 â€“ 8 = 18

dÂ = 3

Now,

A1Â =Â aÂ +Â dÂ = 8 + 3 = 11

A2Â =Â aÂ + 2dÂ = 8 + 2 Ã— 3 = 8 + 6 = 14

A3Â =Â aÂ + 3dÂ = 8 + 3 Ã— 3 = 8 + 9 = 17

A4Â =Â aÂ + 4dÂ = 8 + 4 Ã— 3 = 8 + 12 = 20

A5Â =Â aÂ + 5dÂ = 8 + 5 Ã— 3 = 8 + 15 = 23

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

15. IfÂ is the A.M. betweenÂ aÂ andÂ b, then find the value ofÂ n.

Solution:

The A.M between a and b is given by, (a + b)/2

Then according to the question,

Thus, the value of n is 1.

16. Between 1 and 31,Â mÂ numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7thÂ and (mÂ â€“ 1)thÂ numbers is 5: 9. Find the value ofÂ m.

Solution:

Letâ€™s consider a1, a2, â€¦ amÂ beÂ mÂ numbers such that 1, a1, a2, â€¦ am, 31 is an A.P.

And here,

aÂ = 1,Â bÂ = 31,Â nÂ =Â mÂ + 2

So, 31 = 1 + (mÂ + 2 â€“ 1) (d)

30 = (mÂ + 1)Â d

d = 30/ (m + 1) â€¦â€¦. (1)

Now,

a1Â =Â aÂ +Â d

a2Â =Â aÂ + 2d

a3Â =Â aÂ + 3dÂ â€¦

Hence, a7Â =Â aÂ + 7d

amâ€“1Â =Â aÂ + (mÂ â€“ 1)Â d

According to the question, we have

Therefore, the value of m is 14.

17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30thÂ instalment?

Solution:

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105 and so on as the instalment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And the, A.P. is 100, 105, 110, â€¦

Where, first term,Â aÂ = 100

Common difference,Â dÂ = 5

So, the 30th term in this A.P. will be

A30Â  =Â aÂ + (30 â€“ 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30thÂ instalment will be Rs 245.

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

Solution:

Itâ€™s understood from the question that, the angles of the polygon will form an A.P. with common differenceÂ dÂ = 5Â° and first termÂ a = 120Â°.

And, we know that the sum of all angles of a polygon withÂ nÂ sides is 180Â° (nÂ â€“ 2).

Thus, we can say

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.