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

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 the 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 examinations. Students have to ensure that they practise every problem given in the textbook repeatedly until the concept gets clear.

#### 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, …1999, 2001.

It clearly forms a sequence in A.P.

Where, the first term, *a* = 1

Common difference, *d* = 2

Now,

a + (n -1)d = 2001

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

2n – 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, … 995.

It clearly forms a sequence in A.P.

Where, the first term, *a* = 105

Common difference, *d* = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 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 20th term is –112.**

**Solution:**

Given,

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

Let’s assume *d* be the common difference of the A.P.

So, the A.P. will be 2, 2 + *d*, 2 + 2*d*, 2 + 3*d*, …

Then,

Sum of first five terms = 10 + 10*d*

Sum of next five terms = 10 + 35*d*

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

Therefore, the 20th term of the A.P. is –112.

**4. How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?**

**Solution:**

Let’s consider the sum of *n* terms of the given A.P. as –25.

We known that,

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

where *n* = number of terms, *a* = first term, and *d* = common difference

So here, *a* = –6

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

Thus, we have

**5. In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first pq terms is ½ (pq + 1) where p ≠ q. **

**Solution:**

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

**Solution:**

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 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 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 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 – 1)(-3)

= 25 – 21

= 4

**7. Find the sum to n terms of the A.P., whose kth term is 5k + 1.**

**Solution:**

Given, the *k*th term of the A.P. is 5*k* + 1.

*k*th term = *ak* = *a *+ (*k* – 1)*d*

And,

*a *+ (*k* – 1)*d* = 5*k* + 1

*a* + *kd* – *d* = 5*k* + 1

On comparing the coefficient of *k*, we get *d* = 5

*a *– *d *= 1

*a* – 5 = 1

⇒ *a* = 6

**8. If the sum of n terms of an A.P. is (pn + qn2), where p and q 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 of *n*2 on both sides, we get

d/2 = q

Hence, *d* = 2*q*

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

**9. 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 *a*1, *a*2, and *d*1, *d*2 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 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.**

**Solution:**

Let’s take *a* and *d* 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 first p, q and r terms of an A.P. are a, b and c, respectively.**

**Prove that **

**Solution:**

Let *a*1 and *d* 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 of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).**

**Solution:**

Let’s consider that *a* and *b* 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 of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.**

**Solution:**

Let’s consider *a* and *b* to be the first term and the common difference of the A.P. respectively.

*am* = *a* + (*m* – 1)*d* = 164 … (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:**

Let’s assume A1, A2, A3, A4, and A5 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,

*a *= 8, *b *= 26, *n* = 7

So,

26 = 8 + (7 – 1) *d*

6*d* = 26 – 8 = 18

*d *= 3

Now,

A1 = *a* + *d* = 8 + 3 = 11

A2 = *a* + 2*d* = 8 + 2 × 3 = 8 + 6 = 14

A3 = *a* + 3*d* = 8 + 3 × 3 = 8 + 9 = 17

A4 = *a* + 4*d *= 8 + 4 × 3 = 8 + 12 = 20

A5 = *a* + 5*d* = 8 + 5 × 3 = 8 + 15 = 23

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

**15. If is the A.M. between a and b, then find the value of 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, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5: 9. Find the value of m.**

**Solution:**

Let’s consider a1, a2, … a*m* be *m* numbers such that 1, a1, a2, … a*m*, 31 is an A.P.

And here,

*a* = 1, *b* = 31, *n* = *m* + 2

So, 31 = 1 + (*m* + 2 – 1) (*d*)

30 = (*m* + 1) *d*

d = 30/ (m + 1) ……. (1)

Now,

a1 = *a* + *d*

a2 = *a* + 2*d*

a3 = *a* + 3*d* …

Hence, a7 = *a* + 7*d*

a*m*–1 = *a* + (*m* – 1)* 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 30th 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, …

Where, first term, *a* = 100

Common difference, *d* = 5

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

A30 = *a* + (30 – 1)*d*

= 100 + (29) (5)

= 100 + 145

= 245

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

**18. 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.**

**Solution:**

It’s understood from the question that, the angles of the polygon will form an A.P. with common difference *d* = 5° and first term *a* = 120°.

And, we know that the sum of all angles of a polygon with *n* sides is 180° (*n* – 2).

Thus, we can say

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