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

We are providing NCERT Solutions of Chapter 9 Sequence and Series Exercise 9.2 here that will be useful for every Class 11 students who want to improve their marks in the examination. NCERT Solutions for Class 11 Maths provided here are prepared by Studyrankers are accurate and detailed which will help you in knowing the important points of chapter. 1. Find the sum of odd integers from 1 to 2001.

We have to find the sum 1 + 3 + 5 + ......+2001
Let the n th term be the last term.
a + (n –1)d = 2001
a = 1, d = 2
1 + [n – 1]. 2 = 2001 or 2[n –1]
= 2001 – 1 =2000
n – 1 = 1000, n = 1001
Sum = n/2[2a + (n – 1)d]
= 1001/2 [2.1 + (1001 – 1) × 2]
= 1001 × 1001 = 1002001

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

We have to find the sum
105 + 110 + 115 + ........+ 995
Let 995 = n th term
∴ a + [n – 1]d = 995 or 105 + [n – 1]5 = 995
Dividing by 5,
21 + (n – 1) = 199 or n = 199 – 20 = 179
∴ 105 + 110 + 115 + ........+ 995
= n/2 [2a + (n – 1)d]
= 179/2 [2 × 105 + (179 – 1)5]
= 179/2 [2×105 + 5 × 178] = 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.

Let d be the common difference of A.P.
∴ Sum of first 5 term = n/2[2a + (n – 1)d]
= 5/2[ 2 × 2 + (5 – 1)d] = 5(2 + 2d)
6th term = a + (6 – 1)d = a + 5d
∴ (a + 5d) + (a + 6d) + ………… to 5 terms
Sum of the next 5 terms
= 5/2 [2(a + 5d) + (5 – 1)d]
= 5/2 [2a + 10d + 4d] = 5/2[2a + 14d] = 5(a + 7d)
Sum of first five term
= 1/4 × sum of next 5 terms
5(2 + 2d) = 1/4 × 5 (2 + 7d)
=> 4(2 + 2d) = 2 + 7d ∴ d= -6
∴ T20 = a + (20 – 1)d = 2 + 19 × (-6)
= 2 + (-114) = - 112

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

We have a = –6, d= -(11/2)- (–6)
d = (-11 + 12)/2 = 1/2
Let n be the number of terms so that Sn = –25
Since, Sn = n/2[2a + (n – 1)d]
=> -25 = n/2[2 . -6 + (n -1) × 1/2]
=> -25 = n/2[-12 + (n -1)/2]
=> -25 = n/2[(-24 + n – 1)/2]
=> -25 = n/2[(n – 25)/2] => –100 = n(n – 25)
=> n 2 – 25n + 100 = 0
=> (n – 5)(n – 20) = 0
=> n = 5 or n = 20
Both values of n give the required sum.

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 1/2 (pq +1), where p ≠ q.

Let a be the first term and d be the common difference of an A.P. then
Tp = 1/q => a + (p – 1)d = 1/q ………(i)
Tq = 1/p => a + (q – 1)d = 1/p ……(ii)
Eqn (i) – eqn (ii) gives,
(p – q)d = 1/q – 1/p = (p – q)/pq => d = 1/pq
Putting d = 1/pq in (i), we get
a + (p – 1) 1/pq = 1/q => a + 1/q – 1/pq = 1/q
=> a = 1/pq
Now, Spq = pq/2[2a + (pq – 1)d]
= pq/2 [2/pq + (pq – 1) 1/pq]
= pq/2 [2/pq + 1 – 1/pq] = pq/2 (1/pq + 1)
Spq = 1/2(pq + 1)

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

a = 25, d = 22 – 25 = –3. let n be the no. of terms
Sum = 116; Sum = n/2[2a + (n – 1)d]
116 = n/2 [50 + (n – 1)(-3)]
Or 232 = n[50 – 3n + 3] = n[53 – 3n]
= -3n2 + 53n
=> 3n2 – 53 + 232 = 0
=> (n – 8)(3n – 29) = 0
=> n = 8 or n = 29/3, n ≠ 29/3 ∴ n = 8
∴ Now, T8 = a + (8 – 1)d = 25 + 7 . (-3)
= 25 – 21
∴ Last term = 4

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

Tk =5k = + Putting k = 1, 2
T1 = 5 × 1 + 1 = 5 + 1 = 6;
T2 = 5 × 2 + 1 = 10 + 1 = 11
∴ d = T2 – T1 = 11 – 6 = 5
a = 6, d = 5
Sum of n term = n/2 [2a + (n – 1)d]
= n/2 [2 × 6 + (n – 1)5]
= n/2 [12 + 5n – 5] = (n(5n + 7))/2

8. If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

Let Sn be the sum of n term Sn = pn + qn2
Putting n = 1, 2
S1 = T1 = p. 1 + q. 12 = p + q,
S2 = T1 + T2 = p. 2 + q. 22 = 2p + 4q
T2 = (T1 + T2) – T1 = S2 – S1
= [(2p + 4q) – (p + q)] = p + 3q
d = T2 – T1 = (p + 3q) – (p + q) = 2q
Common difference of the series is 2q.

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.

Let a1, d1 be the first term and common difference of the first A.P. and a2, d2 be the first term and common difference of the second A.P. If S1 and S2 be their sums respectively. Then
S1 = n/2 [2a1 + (n – 1)d1];
S2 = n/2[2a2 + (n – 1)d2]
S1/S2 = (n/2[2a1 + (n – 1)d1])/(n/2[2a2 + (n – 1)d2])
= (2a1 + (n – 1)d2)/(2a2 + (n – 1)d2) ….(i)
S1/S2 = (5n + 4)/(9n + 6) ……(ii)
We have to find
(18th term of first A.P.)/(18th term of second A.P) = (a1 + 17d1)/(a2 + 17d2)
Multiplying the numerator and denominator by 2
(T18 of I AP)/(T18 of II A.P) = (2a1 + 34d1)/(2a2 + 34d2) ……..(iii)
Comparing (i) and (iii)
(n – 1)d1 = 34d1 ∴ n = 34 + 1 = 35
Putting n = 35 in (ii)
(2a1 + (35 – 1)d1)/(2a2 + (35 – 1)d2) = (5 × 35 + 4)/(9 × 35 + 6)
=> (2a1 + 34d1)/(2a2 + 34d2) = (175 + 4)/(315 + 6) = 179/321
(T18 of Ist A.P)/T18 of IIInd A.P) = 179/321.

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.

Let a be the first term and d be the common difference of A.P.
Sum of first p terms = p/2 [2a + (p – 1)d] … (i)
Sum of first q terms = q/2 [2a + (q – 1)d] … (ii)
Equating (i) & (ii)
p/2[2a + (p – 1)d] = q/2[2a + (q – 1)d]
Transposing the term of R.H.S to L.H.S
or 2a(p – 1) + p(p – 1) – q(q – 1)d = 0
=> 2a(p – 1) + [(p2 – q2 – (p – q)d] = 0
Or 2a(p – 1) + (p – q)[(p + q) – d] = 0
=> (p – q)[2a + (p + q – 1)d] = 0
=> 2a + (p + q – 1)d = 0 …….(iii)
(p ≠ q)
Sum of first (p + q) term
= (p + q)/2 [2a + (p + q – 1)d] = (p + q)/2 × 0 = 0
∴ 2a + (p + q – 1)d = 0 [from (iii)]

11. Sum of the first p, q and r terms of an A.P. are a, b and c respectively.
Prove that
a/p (q – r) + b/q (r – p) + c/r (p – q) = 0

Let A be the first term and D the common difference of A.P.
Sp = a
=> p/2[2A + (p – 1)D] = a
=> 1/2[2A + (p – 1)D] = a/p
=> A + 1/2 (p – 1) D = a/p ……(i)
Similarly, A + 1/2 (q – 1)D = b/q ……(ii)
And A + 1/2 (r – 1)D = c/r ………(iii)
Multiplying (i) by (q – r) (ii) by (r – p) (iii) by (p – q) and adding, we get
A(q – r + r – p + p – q) + 1/2 [(p – 1)(q – r) + (q -1)(r – p) + (r – 1)(p – q)]
= a/p (q – r) + b/q (r – p) + c/r (p – q)
=> A(0) + 1/2 (0)D = a/p (q – r) + b/q (r – p) + c/r (p – q)
Hence, a/p (q – r) + b/q (r – p) + c/r (p – q) = 0.

12. The ratio of the sums of m and n terms of an A.P. is m2: n 2. Show that the ratio of mth and n th term is (2m – 1): (2n – 1).

Let a be the first term and d be the common difference of A.P.
then (m/2 [2a + (m – 1)d]/(n/2[2a + (n – 1)d] = m2/n2
=> (2a + (m – 1)d)/(2a + (n – 1)d) = m/n …..(i)
For the ratio of mth term to the n th term, to find the value of
Tm/Tn = (a + (m – 1)d)/(a + (n – 1)d) = (2a + (2m – 2)d)/(2a + (2n – 2)d) …(ii)
LH.S of eqn (i) will be identical with (ii) if (m –1) replaces (2m – 2) and (n – 1) replaces (2n – 2)
So, replacing m by 2m – 1 and n by 2n – 1 on both sides of (i), we get
(2a + (2m – 2)d)/(2a + (2n – 2)d) = (2m – 1)/(2n – 1)

13. If the sum of n terms of an A.P. is 3n 2 + 5n and its mth term is 164, find the value of m.

Let the sum of n term is denoted by Sn
∴ Sn = 3n 2 + 5n
Put n = 1, 2. T1 = S1 = 3.12 + 5.1 = 3 + 5 = 8;
S2 = T1 + T2 = 3. 22 + 5. 2 = 12 + 10 = 22
∴ T2 = S2 – S1 = 22 – 8 = 14
∴ Common difference d = T2 – T1 = 14 – 8 = 6
a = 8, d = 6
mth term = a + (m – 1)d = 164 => 8 + (m – 1) . 6 = 164
6m + 2 = 164 => 6m = 164 – 2 = 162
∴ m = 162/6 = 27

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

Let the five numbers be denoted by T2, T3, T4, T5, T6 and 26 are in A.P.
26 is the 7th term. If d is the common difference
Then T7 = a + (7 – 1) d
26 = 8 + 6d = 6d = 26 – 8 =18
d = 3
T2 = a + d = 8 + 3 = 11
T3 = a + 2d = 8 + 6 = 14
T4 = a + 3d = 8 + 9 = 17
T5 = a + 4d = 8 + 12 = 20
T6 = a + 5d = 8 + 15 = 23
Required five numbers are 11, 14, 17, 20, 23.

15. If (an + bn)/(an-1 + bn – 1) is the A.M between a and b, then find the value of n.

A.M between a and b = (a + b)/2
∴ (an + bn)/(an – 1 + b n-1) = (a + b)/2
2an + 2bn = an + abn-1 + an-1b + bn
=> anan-1b – abn-1 + bn = 0
=> an-1(a – b)-bn-1(a- b) = 0
=> (a – b)(an – bn-1) + bn = 0
[a ≠ 0]
=> an-1 – bn-1 = 0 => an-1 = bn-1
(a/b)n-1 = 1 = (a/b)0 => n -1 = 0 => n = 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.

Let A1, A2, .........Am be the m arithmetic means each inserted between 1 and 31.
Total number of terms = m + 2
If d is the common difference, then
=> Tm+2= 1 + (m + 2 – 1)d = 31 => (m + 1) d = 30 => d = 30/(m + 1)
A7 = T8 = a + 7d
= 1 + 7 × 30/(m + 1) = (m + 1 + 210)/(m + 1) = (m + 211)/(m + 1)
Am-1 = Tm
= a + (m – 1)d
= 1 + (m – 1)× 30/(m + 1) = (m + 1 + 30m – 30)/(m + 1)
= (31m – 29)/(m + 1)
Now, A7/Am-1 = (m + 211)/(31m – 29) = 5/9
=> 9m + 1899 = 155m – 145 => 146m = 2044
=> m = 2044/146 = 14 => m = 14

17. A man starts repaying a loan as first installment of ₹ 100. If he increase the installment by ₹ 5 every month, what amount he will pay in the 30th installment?

The instalments form an A.P. whose first difference is ₹ 100 and the common difference is ₹ 5.
30th instalment = T30 = a + (30 – 1)d
= 100 + 29 × 5
= 100 + 145 = 245
Thus, the 30th instalment is of ₹ 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.