Practicing Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(c) is the ultimate need for students who intend to score good marks in examinations.
S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(c)
Question 1.
Find the sum :
(i) 10 terms of 5 + 8 + 11 +….;
(ii) 18 terms of 57 + 49 + 41 +….;
(iii) n terms of 4 + 7 + 10 +….;
(iv) 24 terms and n terms of 2\(\frac { 1 }{ 2 }\), 3\(\frac { 1 }{ 3 }\), 4\(\frac { 1 }{ 6 }\), 5 ….;
Solution:
(i) Clearly given series forms A.P with first term a = 5 and common difference d = 8 – 5 = 3; n = 10
We know that Sn = \(\frac { n }{ 2 }\)[2a + (n – 1) d]
∴ S10 = \(\frac { 10 }{ 2 }\) [2 × 5 + (10 – 1) 3]
= 5 [10 + 27] = 185
(ii) Clearly given series forms A.P with first term a = 57 and common difference
d = 49 – 57 = – 8 ; n = 18
We know that Sn = \(\frac { n }{ 2 }\)[2a + (n – 1) d]
∴ S18 = \(\frac { 18 }{ 2 }\) [2 × 57 + (18 – 1) (-8)]
= 9[114 – 136] = 9 × (-22) = – 198
(iii) Clearly given series forms A.P with first term a = 2\(\frac { 1 }{ 2 }\)
and common difference d = 3\(\frac { 1 }{ 2 }\) – 2\(\frac { 1 }{ 2 }\)
= \(\frac { 10 }{ 3 }\) – \(\frac { 5 }{ 2 }\) = \(\frac { 5 }{ 6 }\)
We know that Sn = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
∴ S24 = \(\frac { 24 }{ 2 }\) [2 × \(\frac { 5 }{ 2 }\) + (24 – 1)\(\frac { 5 }{ 6 }\)]
= 12[5 + \(\frac { 115 }{ 6 }\)] = \(\frac{12 \times 145}{6}\) = 290
and Sn = \(\frac { n }{ 2 }\) [2 × \(\frac { 5 }{ 2 }\) + (n – 1)\(\frac { 5 }{ 6 }\)]
= \(\frac { n }{ 2 }\)[5 + (n – 1)\(\frac { 5 }{ 6 }\)] = \(\frac { n }{ 12 }\) (5n + 25)
= \(\frac { 5n }{ 12 }\)(n + 5)
(v) Given series forms an A.P with first term a = 101 and common difference
d = 99 – 101 = -2
Since l = 47
⇒ 47 = Tn = a +(n – 1) d
⇒ 47 = 101 + (n – 1)(- 2)
⇒ (n – 1)(- 2) = -54
⇒ n – 1 = 27
⇒ n = 28
We know that Sn = \(\frac { n }{ 2 }\)[a + l]
⇒ S28 = \(\frac { 28 }{ 2 }\) [101 + 47] = 14 × 148 = 2072
Question 2.
Find the sum of all the numbers between 100 and 200 which are divisible by 7 .
Solution:
Numbers between 100 and 200 which are divisible by 7 are 105, 112, ……, 196
Clearly given progression forms an A.P with first term a = 105 and common difference = d = 112 – 105 = 7
Let n be the no. of terms of given A.P
s.t Tn = 196
⇒ 196 = a + (n – 1) d
⇒ 196 = 105 + (n – 1) 7
⇒ 91 = (n – 1) 7 ⇒ n – 1 = 13
⇒ n = 14
We know that, Sn = \(\frac { n }{ 2 }\)[a + l]
∴ S14 = \(\frac { 14 }{ 2 }\)[105 + 196] = 7 × 301 = 2107
Question 3.
The sum of a series of terms in A.P. is 128. If the first term is 2 and the last term is 14 , find the common difference.
Solution:
Let a be the first term and be the common difference of given A.P
Then a = 2; d = 14 and Sn = 128
We know that, Sn = \(\frac { n }{ 2 }\)(a + l)
⇒ 128 = \(\frac { n }{ 2 }\)(2 + 14) ⇒ n = \(\frac { 128 }{ 8 }\) = 16
Since last term = 14 ⇒ 14 = a + (n – 1) d
⇒ 14 = 2 + (16 – 1) d ⇒ 12 = 15d
⇒ d = \(\frac { 4 }{ 5 }\)
Question 4.
The sum of 30 terms of a series in A.P., whose last term is 98 , is 1635 . Find the first term and the common difference.
Solution:
Let a be the first term and d be the common difference of given A.P.
Here n = 30 s.t S30 = 1635 and l = 98
We know that Sn = \(\frac { n }{ 2 }\)(a + l)
⇒ S30 = \(\frac { 30 }{ 2 }\)(a + 98)
⇒ 1635 = 15(a + 98)
⇒ a + 98 = 109 ⇒ a =11
Further l = Tn = a + (n – 1) d = 98
⇒ 11 +(30 – 1) d = 98
⇒ 29 d = 87
⇒ d = 3
Question 5.
If the sums of the first 8 and 19 terms of an A.P. are 64 and 361 respectively, find (i) the common difference and (ii) the sum of n terms of the series.
Solution:
Let a be the first term and d be the common difference of given A.P.
given S8 = 64 ⇒ \(\frac { 8 }{ 2 }\)[2a + (8 – 1)d] = 64
⇒ 2a + 7d = 16 …(1) [∵ Sn = \(\frac { n }{ 2 }\) [2a + (n – 1)d]]
Also S19 = 361
⇒ \(\frac { 19 }{ 2 }\) [2a + (19 – 1) d] = 361
⇒ 2a + 18d = 38 …(2)
eqn. (2) – eqn. (1) gives;
11d = 22 ⇒ d = 2
∴ from (1); 2a + 14 = 16 ⇒ a = 1
Since Sn = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
= \(\frac { n }{ 2 }\) [2 × 1 + (n – 1) 2]
= \(\frac { n }{ 2 }\) [2 + 2n – 2] = n2
Question 6.
Find the number of terms of the series 21, 18, 15, 12, ……. which must be taken to give a sum of zero.
Solution:
Clearly the given series forms A.P with first term a = 21 and common difference d = -3 Let n be the required no. of terms of given A.Ps and Sn = 0
We know that
Sn = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
⇒ Sn = 0 ⇒ 0 = 2a + (n – 1) d
⇒ 0 = 2 × 21 + (n – 1) (-3)
⇒ – 42 = (n – 1) (- 3)
⇒ n – 1 = 14
⇒ n = 15
Hence the required number of terms of given A.P be 15 .
Question 7.
The sum of n terms of a series is (n2 + 2n) for all values of n. Find the first 3 terms of the series.
Solution:
Given Sn = n2 + 2n
∴ Sn – 1 = (n – 1)2 + 2(n – 1)
Thus, Tn = Sn – Sn – 1
= n2 + 2n – (n – 1)2 – 2(n – 1)
⇒ Tn = n2 + 2n – (n2 – 2n + 1) – (2n – 2)
⇒ Tn = 2n + 1
∴ T1 = 2 × 1 + 1 = 3
T2 = 2 × 2 + 1 = 5
and T3 = 2 × 3 + 1 = 7
Hence the first three terms of given A.P be 3, 5 and 7 .
Question 8.
The third term of an arithmetical progression is 7, and the seventh term is 2 more than 3 times the third term. Find the first term, the common difference and the sum of the first 20 terms.
Solution:
Let a be the first term and d be the common difference of given A.P.
given T3 = 7 ⇒ a + 2d = 7 …(1)
and T7 = 2 + 3T3 ⇒ T7 = 2 + 3 × 7
⇒ a + 6d = 23 …(2)
eqn. (2) – eqn. (1) gives ;
4d = 16 ⇒ d = 4
∴ from (1); a + 8 = 7 ⇒ a = – 1
We know that Sn = \(\frac { n }{ 2 }\)[2 a+(n-1) d]
∴ S20 = \(\frac { 20 }{ 2 }\)[2(- 1) + (20 – 1) 4]
= 10[- 2 + 76]
⇒ S20 = 740
Question 9.
The interior angles of a polygon are in arithmetic progression. The smallest angle is 52° and the common difference is 8°. Find the number of sides of the polygon.
Solution:
Let a be the smallest angle ∴ a = 52°
and common difference = d = 8°
Let n be the required no. of sides of polygon Then Sn = sum of all interior angles of polygon with side n
⇒ \(\frac { n }{ 2 }\) [2 × 52° + (n – 1) 8°] = (n – 2) × 180°
⇒ n[52° + (n – 1) 4°] = (n – 2) 180°
⇒ n[52° + 4n – 4°] = (n – 2) 180°
⇒ n (4n + 48) = (n – 2) 180°
⇒ n(n + 12) = (n – 2) 45
⇒ n2 – 33n + 90 = 0
⇒ (n – 3) (n – 30) = 0 ⇒ n = 30, 3
When n = 30; T30 = 50° + 29 × 8°
= 52° + 232° = 284°
which is not possible. Since any interior angle of polygon cannot be more than 180°.
∴ n = 3
Hence the required number of sides of the polygon be 3 .
Question 10.
Determine the sum of first 35 terms of an A.P. if t2 = 1 and t7 = 22.
Solution:
Let a be the first term and d be the common difference of given A.P.
given t2 = 1 ⇒ a + d = 1 ….(1)
and t7 = 22 ⇒ a + 6d = 22 …(2)
eqn. (2) – eqn. (1) gives;
5d = 21 ⇒ d = \(\frac { 21 }{ 5 }\)
∴ from (1); a = 1 –\(\frac { 21 }{ 5 }\) = \(\frac { -16 }{ 5 }\)
We know that Sn = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
∴ S35 = \(\left[2\left(-\frac{16}{5}\right)+34 \times \frac{21}{5}\right]\)
= \(\frac { 35 }{ 2 }\) \(\left[\frac{-32}{5}+\frac{714}{5}\right]\) = \(\frac { 35 }{ 2 }\) × \(\frac { 682 }{ 5 }\)
= 7 × 341 = 2387
Question 11.
Find the sum of all natural numbers between 100 and 1000 which are multiples of 5 .
Solution:
All natural numbers between 100 and 1000 which are multiples of 5 are 105, 110, 115, …, 995
it clearly forms A.P with first term a = 105 and d = 110 – 105 = 5; l = 995
Let n be the required no. of terms of given A.P.
∴ 995 = a + (n – 1) d
⇒ 995 = 105 + (n – 1) 5
⇒ 995 – 105 = (n – 1) 5 ⇒ \(\frac { 890 }{ 5 }\) =n – 1
⇒ n – 1 = 178 ⇒ n = 179
We know that Sn = \(\frac { n }{ 2 }\) = [a + l]
∴ S179 = [105 + 995] = \(\frac { 179 }{ 2 }\) × 1100 = 98450
Question 12.
How many terms of the A.P. 1, 4, 7, ….are needed to give the sum 715 ?
Solution:
Let a be the first term and d be the common difference of given A.P.
Then a = 1, d = 4 – 1 = 3
Let n be the required no. of terms of given A.P s.t Sn = 715
⇒ 715 = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
⇒ 715 = \(\frac { n }{ 2 }\) [2 × 1 + (n – 1) 3]
⇒ 1430 = n[2 + 3n – 3]
⇒ 1430 = n (3n – 1)
⇒ 3n2 – n – 1430 = 0
⇒ n = \(\frac{1 \pm \sqrt{1+4 \times 3 \times 1430}}{6}\)
= \(\frac{1 \pm \sqrt{17161}}{6}\) = \(\frac{1 \pm 131}{6}\)
⇒ n = 22, –\(\frac { 65 }{ 3 }\)
But n ∈ N
∴ n = 22
Hence the required no. of terms of given A.P be 22 .
Question 13.
Find the rth term of an A.P., sum of whose first n terms is 2n + 3n2.
Solution:
Given Sn = 2n + 3n2
∴ Sn – 1 = 2(n – 1) + 3(n – 1)2
Thus Tn = Sn – Sn – 1
= [2n + 3n2] – [2(n – 1) + 3(n – 1)2]
= 2n + 3n2 – [2n – 2 + 3n2 – 6n + 3]
= 2n + 3n2 – [3n2 – 4n + 1]
Tn = 6n – 1
∴ Tr =6r – 1
Question 14.
In an arithmetical progression, the sum of p terms is m and the sum of q terms is also m. Find the sum of (p + q) terms.
Solution:
Let a be the first term and d be the common difference of given A.P.
It is given that Sp = m
⇒ \(\frac { p }{ 2 }\) [2a + (p – 1)d] = m
⇒ ap = \(\frac { p }{ 2 }\) (p – 1)d = m …(1)
and Sq = m ⇒ \(\frac { q }{ 2 }\)[2a + (q – 1) d] = m
⇒ aq + \(\frac { q }{ 2 }\) (q – 1)d = m …(2)
eqn. (1) – eqn. (2) gives ;
a(p – q) + \(\frac { d }{ 2 }\) [p2 – p – q2 + q] = 0
⇒ a(p – q) + \(\frac { d }{ 2 }\) [(p – q) (p + q) – 1 (p – q)] = 0
⇒ (p – q) a + \(\frac { d }{ 2 }\) (p – q) [p + q – 1] = 0
⇒ a + \(\frac { d }{ 2 }\) (p + q – 1) = 0 …(3) [∵ p – q ≠ 0]
∴ Sp + q = \(\frac { p + q }{ 2 }\) [2a + (p + q – 1)d]
= (p + q) [a + (p + q – 1)\(\frac { d }{ 2 }\)]
= (p + q) × 0 = 0
Question 15.
The sum of the first fifteen terms of an arithmetical progression is 105 and the sum of the next fifteen terms is 780 . Find the first three terms of the arithmetical progression.
Solution:
Let a be the first term and d be the common difference of given A.P.
given S15 = 105
⇒ \(\frac { 15 }{ 2 }\) [2a + (15 – 1) d] = 105
⇒ 2a + 14d = 14 …(1)
Also, S30 – S15 = 780
⇒ S30 = 780 + 105 = 885
⇒ \(\frac { 30 }{ 2 }\) [2a + (30 – 1)d] = 885
⇒ 2a + 29d = \(\frac { 885 }{ 15 }\) = 59 …(2)
eqn. (2) – eqn. (1) gives;
15 d = 45 ⇒ d = 3
∴ from (1); 2a + 42 = 14 ⇒ a = -14
Thus the first three terms of A.P are
a, a + d, a + 2d
i.e. – 14, – 14 + 3, – 14 + 6
i.e. – 14, – 11, – 8
Question 16.
16. The sum of the first six terms of an arithmetic progression is 42 . The ratio of the 10th term to the 30 th term of the A.P. is \(\frac { 1 }{ 3 }\). Calculate the first term and the 13th term.
Solution:
Let a be the first term and d be the common difference of an A.P.
∴ S6 = 42
⇒ \(\frac { 6 }{ 2 }\) [2a + (6 – 1)d] = 42
⇒ 2a + 5d = 14 …(1)
Also \(\frac{T_{10}}{T_{30}}\) = \(\frac { 1 }{ 3 }\) ⇒ 3T10 T30
⇒ 3(a + 9d) = a + 29d
⇒ 2a = 2d ⇒ a = d
∴ from (1); 7a = 14
∴ a = 2 = d
∴ T13 = a + 12d = 2 + 12 × 2 = 26
Question 17.
A sum of ₹ 6240 is paid off in 30 instalments, such that each instalment is ₹ 10 more than the preceeding instalment. Calculate the value of the first instalment.
Solution:
Let the value of first instalment be ₹ a.
Then the sequence of instalments is given as under :
a, a + 10, a + 20, a + 30, …… 30 terms
Clearly it form an A.P with common difference d = 10
Also it is given that S30 = 6240
⇒ 6240 = \(\frac { 30 }{ 2 }\)[2a + (30 – 1) 10]
[∵ Sn = \(\frac { n }{ 2 }\)[2 a+(n-1) d]]
⇒ 6240 = 15(2a + 290)
⇒ 2a + 290 = 416
⇒ 2a = 416 – 290 = 126
⇒ a = 63
Question 18.
The nth term of an A.P. is p and the sum of the first n term is s. Prove that the first term is \(\frac{2 s-p n}{n}\).
Solution:
Let a be the first term and d be the common difference of given A.P.
given Tn = last term = p
Also Sn = \(\frac { n }{ 2 }\) [a + l] ⇒ s = \(\frac { n }{ 2 }\) [a + p]
⇒ \(\frac { 2s }{ n }\) – p = a
⇒ a = \(\frac{2 s-n p}{n}\)
Question 19.
The sum of the first n terms of the arithmetical progression 3, 5\(\frac { 1 }{ 2 }\), 8, …. is equal to the 2nth term of the arithmetical progression 16\(\frac { 1 }{ 2 }\), 28\(\frac { 1 }{ 2 }\), 40\(\frac { 1 }{ 2 }\). Calculate the value of n.
Solution:
Given first series be 3, 5\(\frac { 1 }{ 2 }\), 8, ….
it clearly forms A.P with first term a = 3 and common difference d = \(\frac { 11 }{ 2 }\) – 3 = \(\frac { 5 }{ 2 }\)
We know that Sn = \(\frac { n }{ 2 }\)[2 a +(n – 1) d]
∴ Sn = \(\frac { n }{ 2 }\) \(\left[2 \times 3+(n-1) \frac{5}{2}\right]\)
= \(\frac { n }{ 2 }\) \(\left[6+(n-1) \frac{5}{2}\right]\)
⇒ Sn = \(\frac { n }{ 4 }\) [12 + 5n – 5] = \(\frac { n }{ 4 }\) [5n + 7] …(1)
Also, given second series be,
16\(\frac { 1 }{ 2 }\), 28\(\frac { 1 }{ 2 }\), 40\(\frac { 1 }{ 2 }\), ….
it clearly forms A.P with first term A = \(\frac { 33 }{ 2 }\) and common diff. D =\(\frac { 57 }{ 2 }\) – \(\frac { 33 }{ 2 }\) = 12
∴ \(\mathrm{T}_{2 n}^{\prime}=[\mathrm{A}+(2 n-1) \mathrm{D}]\)
= \(\left[\frac{33}{2}+(2 n-1) 12\right]\)
= \(\left[24 n+\frac{9}{2}\right]\)
According to given condition, we have
Sn = \(\mathrm{T}_{2 n}^{\prime}\)
⇒ \(\frac { n }{ 4 }\)(5n + 7) = \(\left[\frac{33}{2}+(2 n-1) 12\right]\)
⇒ \(\frac { n }{ 4 }\)(5n + 7) = \(\left[24 n+\frac{9}{2}\right]\)
⇒ n(5n + 7) = 96n + 18
⇒ 5n2 – 89n – 18 = 0
∴ n = \(\frac{89 \pm \sqrt{7921+360}}{10}\)
⇒ n = \(\frac{89 \pm 91}{10}\) = 18, – \(\frac { 1 }{ 5 }\)
Since n ∈ N
∴ n = 18
Question 20.
If the sum of the first 4 terms of an arithmetic progression is p, the sum of the first 8 terms is q and the sum of the first 12 terms is n express 3p + r in terms of q.
Solution:
Let a be the first term and d be the common difference of given A.P.
It is given that S4 = p
⇒ \(\frac { 4 }{ 2 }\) [2a + (4 + 1)d] = p
⇒ 4a + 6d = p …(1)
Also, S8 = q ⇒ \(\frac { 8 }{ 2 }\)[2a + (8 – 1) d] = q
⇒ 8a + 28d = q …(2)
Also, S12 = r ⇒ \(\frac {12 }{ 2 }\)[2a + (12 – 1) d] = r
⇒ 12r + 66d = r …(3)
∴ 3p + r = 3(4a + 6d) + 12a + 66d
[using eqn. (1) and (3)]
= 24a + 84d = 3(8a + 28d) = 3q
Question 21.
The last term of an A.P. 2, 5, 8, 11, …. is x. The sum of the terms of the A.P. is 155 . Find the value of x.
Solution:
Clearly given A.P is having first term a = 2
and common difference = d = 5 – 2 = 3
given last term = l = x = a + (n + 1) d
∴ x = 2 + (n – 1) 3
⇒ x = 3n – 1 ⇒ n = \(\frac{x+1}{3}\)
Also, given Sn = 155
⇒ 155 = \(\frac{n}{2}\) [2a + (n + 1)d]
⇒ 155 = \(\left(\frac{x+1}{6}\right)\left[4+\left(\frac{x+1}{3}-1\right) 3\right]\)
⇒ 155 = \(\left(\frac{x+1}{6}\right)\) [4 + (x + 1 – 3)]
⇒ 155 = \(\frac{(x+1)}{6}\) (x + 2)
⇒ x2 + 3x – 928 = 0
⇒ x = \(\frac{-3 \pm \sqrt{9+4 \times 928}}{2}\) = \(\frac{-3 \pm 61}{2}\)
⇒ x = 29, – 32
Since last term is x
∴ x = 29 as x ∈ N
Question 22.
A gentleman buys every year Banks’ certificates of value exceeding the last year’s purchase by ₹ 25. After 20 years he finds that the total value of the certificates purchased by him is ₹ 7,250. Find the value of the certificates purchased by him (i) in the 1st year (ii) in the 13th year.
Solution:
Let ₹a be the value of certificates purchased by him.
Value of purchasing certificates every year forms a sequence is under :
₹ a, ₹(a + 25), ₹(a + 50),…. 20 terms, it clearly forms A.P with first term = a
and common difference d = 25
Also given S20 = ₹ 7250
⇒ 7250 = \(\frac{20}{2}\) [2 a+(20-1) 25]
⇒ 7250 = 10[2a + 475]
⇒ 725 – 475 = 2a
⇒ a = 125
Hence the value of certificates purchased by him in first year be ₹ 125 .
Value of certificates purchased by him in 13th year = T13 =a + 12d
= (125 + 12 × 25) = ₹ 425
Question 23.
If the sums of the first n terms of two A.P.’s are in the ratio 7n – 5; 5n + 17; show that the 6th terms of the two series are equal.
Solution:
Let a and A be the first terms of two A.P.’s and d, D be their common differences.
Then \(\frac{\mathrm{S}_n}{\mathrm{~S}_n^{\prime}}\) = \(\frac{7 n-5}{5 n+17}\)
⇒ \(\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 \mathrm{~A}+(n-1) \mathrm{D}]}\) = \(\frac{7 n-5}{3 n+17}\)
⇒ \(\frac{a+\left(\frac{n-1}{2}\right) d}{\mathrm{~A}+\left(\frac{n-1}{2}\right) \mathrm{D}}\) = \(\frac{7 n-5}{5 n+17}\) …(1)
For ratio of 6th terms, putting \(\frac{n-1}{2}\) = 5
i.e. n = 11 in eqn. (1); we have
\(\frac{a+5 d}{A+5 D}\) = \(\frac{7 \times 11-5}{5 \times 11+17}\) = \(\frac{72}{72}\) = 1
⇒ \(\frac{\mathrm{T}_6}{\mathrm{~T}_6^{\prime}}\) = 1 ⇒ T6 = T ‘6
Thus 6th term of both series are equal.
Question 24.
(i) If the ratio of the sum of m terms and n terms of an A.P. be m2 : n2, prove that the ratio of its mth and nth terms is (2m – 1) : (2n – 1).
(ii) Let a1, a2, a3, …… be terms of an A.P.
if \(\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\ldots+a_q}\) = \(\frac{p^2}{q^2}\), (p ≠ q)
then find \(\frac{a_6}{a_{21}}\).
Solution:
(i) Let the two A.P’s are
a, a + d, a + 2d, ……….
and A, A + D, A + 2D, ………..
We have given \(\frac{\mathrm{S}_m}{\mathrm{~S}_n}\) = \(\frac{m^2}{n^2}\)
⇒ \(\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) \mathrm{D}]}\) = \(\frac{m^2}{n^2}\)
⇒ \(\frac{2 a+(m-1) d}{2 a+(n-1) d}\) = \(\frac{m}{n}\)
⇒ \(\frac{a+\left(\frac{m-1}{2}\right) d}{a+\left(\frac{n-1}{2}\right) d}\) = \(\frac{m}{n}\) …(1)
For \(\frac{\mathrm{T}_m}{\mathrm{~T}_n}\), replacing m by 2m – 1 and n
by 2m – 1 in eqn. (1); we have
\(\frac{\mathrm{T}_m}{\mathrm{~T}_n}\) = \(\frac{a+(m-1) d}{a+(n-1) d}\) = \(\frac{2 m-1}{2 n-1}\)
Aliter : Let the two A.P’s are
a, a + d, a + 2d, …….
and A,A + D, A + 2D,…
it is given that \(\frac{\mathrm{S}_m}{\mathrm{~S}_n}\) = \(\frac{m^2}{n^2}\)
⇒ \(\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}\) = \(\frac{m^2}{n^2}\)
⇒ \(\frac{2 a+(m-1) d}{2 a+(n-1) d}\) = \(\frac{m}{n}\)
⇒ 2an + n(m – 1)d = 2am + m(n – 1)d
⇒ 2a(m – n) + d(mn – m – nm + n) = 0
⇒ [2a(m – n) – d(m – n)] = 0
⇒ (m – n)(2a – d) = 0
⇒ 2a = d [∵ m ≠ n]
∴ \(\frac{\mathrm{T}_m}{\mathrm{~T}_n}\) = \(\frac{a+(m-1) d}{a+(n-1) d}\)
= \(\frac{a+(m-1) 2 a}{a+(n-1) 2 a}\) = \(\frac{2 m-1}{2 n-1}\)
(ii) Since a1, a2, a3, ….. are in A.P. Let d be the common difference of given A.P.
Given \(\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\ldots .+a_q}\) = \(\frac{p^2}{q^2}\)
⇒ \(\frac{\text { sum of first } p \text { terms }}{\text { sum of first } q \text { terms }}\) = \(\frac{p^2}{q^2}\)
⇒ \(\frac{\frac{p}{2}\left[2 a_1+(p-1) d\right]}{\frac{q}{2}\left[2 a_1+(q-1) d\right]}\) = \(\frac{p^2}{q^2}\)
⇒ \(\frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}\) = \(\frac{p}{q}\)
⇒ 2a1q (p – 1) d = 2a1p + p (q – 1)d
⇒ 2a1(p – q) + d(pq – p – pq + q) = 0
⇒ 2a1 (p – q) + d (-p + q) = 0
⇒ (p – q) (2a1 – d) = 0
⇒ 2a1 = d [∵ p ≠ q]
Now \(\frac{a_6}{a_{21}}\) = \(\frac{a_1+5 d}{a_1+20 d}\) = \(\frac{a_1+5 \times 2 a_1}{a_1+20 \times 2 a_1}\) = \(\frac{11 a_1}{41 a_1}\) = \(\frac{11}{41}\)
Question 25.
If the sums of n, 2n, 3n terms of an A.S are S1, S2, S3 respectively, prove that S3 = 3(S2 S1)
Solution:
Let a be the first term and d be the common difference of given A.P respectively.
S1 = Sum of n terms
= \(\frac{n}{2}\)[2a + (n – 1) d] …(1)
S2 = Sum of first 2n terms
= \(\frac{2n}{2}\)[2a + (2n-1) d] …(2)
S3 = Sum of first 3n terms
= \(\frac{3n}{2}\) [2a + (3n – 1)d] …(3)
∴R.H.S = 3 (S2 – S1)
Question 26.
If the sum of p terms of an A.S is q and the sum of q terms is p, show that the sum of (p + q) terms is -(p + q).
Solution:
Let a be the first term and d be the common difference of given A.P.
Given Sp = q ⇒ \(\frac{p}{2}\) [2a + (p – 1)d] = q
⇒ ap + \(\frac{p}{2}\) (p – 1) d = q …(1)
and Sq = p ⇒ \(\frac{q}{2}\) [2a + (q – 1)d] = p
⇒ aq + \(\frac{q}{2}\) (q – 1) d = p …(2)
eqn. (1) – eqn. (2) gives ;
a(p – q) + \(\frac{d}{2}\) [p2 – p – q2 + q] = q + p
⇒ (p – q) \(\left[a+\frac{d}{2}(p+q-1)\right]\) = q + p
⇒ 2a + d(p + q – 1) = – 2 …(3) [∵ p ≠ q]
Thus Sp+q = \(\frac{p+q}{2}[2 a+(p+q-1) d]\)
= \(\left(\frac{p+q}{2}\right)\) (-2)
= – (p + q)
Question 27.
The ratio between the sum of n terms of two A.P.’s is (7n + 1) : (4n + 27). Find the ratio of their 11 th terms.
Solution:
Let the two A.P.’s are a, a + d, a + 2 d, …… and A, A + D, A + 2 D, ….
Given \(\frac{\mathrm{S}_n}{\mathrm{~S}_n^{\prime}}\) = \(\frac{7 n+1}{4 n+27}\)
⇒ \(\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 \mathrm{~A}+(n-1) \mathrm{D}]}\) = \(\frac{7 n+1}{4 n+27}\)
⇒ \(\frac{a+\left(\frac{n-1}{2}\right) d}{\mathrm{~A}+\left(\frac{n-1}{2}\right) \mathrm{D}}=\frac{7 n+1}{4 n+27}\) …(1)
For ratio of 11th terms i.e. \(\frac{\mathrm{T}_{11}}{\mathrm{~T}_{11}^{\prime}}\);
We put \(\frac{n-1}{2}\) = 10 ⇒ n = 21 in eqn. (1);
we have
\(\frac{a+10 d}{a+10 \mathrm{D}}\) = \(\frac{\mathrm{T}_{11}}{\mathrm{~T}_{11}^{\prime}}\) = \(\frac{7 \times 21+1}{4 \times 21+27}\) = \(\frac{148}{111}\)