Accessing Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(b) can be a valuable tool for students seeking extra practice.
S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(b)
Question 1.
Write the first six terms of an A.P. in which
(i) a = 5, d = 4;
(ii) a = 98, d = -3;
(iii) a = 7\(\frac { 1 }{ 2 }\),
d = 1\(\frac { 1 }{ 2 }\);
(iv) a = x, d = 3x + 2.
Solution:
(i) Given a = 5, d = 4
We know that Tn = a + (n – 1)d
⇒ Tn = 5 + (n – 1) 4 = 4n + 1
∴ T1 = 5;
T2 = 4 × 2 + 1 = 9;
T3 = 4 × 3 + 1 = 13;
T4 = 4 × 4 + 1 = 17;
T5 = 4 × 5 + 1 = 21;
T6 = 4 × 6 + 1 = 25
Hence the required six terms of A.P are ; 5, 9, 13, 17, 21 and 25.
(ii) Given a = 98 and d = – 3
We know that Tn = a + (n – 1) d
∴ Tn = 98 + (n – 1)(- 3) = – 3n + 101
∴ T1 = 98 ; T2 = – 3 × 2 + 101 = 95;
T3 = – 3 × 3 + 101 = 92;
T4 = – 3 × 4 + 101 = 89;
T5 = – 3 × 5 + 101 = 86;
T6 = – 3 × 6 + 101 = 83
Hence, the required first six terms of an A.P. are; 98, 95, 92, 89, 86 and 83.
(iii) Given a = 7\(\frac { 1 }{ 2 }\) = \(\frac { 15 }{ 2 }\) and d = 1\(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 2 }\)
We know that,
Tn = a + (n – 1) d
= \(\frac { 15 }{ 2 }\) + ( n – 1)\(\frac { 3 }{ 2 }\) = \(\frac{3 n+12}{2}\)
∴ T1 = a = 7\(\frac { 1 }{ 2 }\);
T2 = \(\frac{3 \times 2+12}{2}\) = 9;
T3 = \(\frac{9+12}{2}\) = 10\(\frac { 1 }{ 2 }\);
T4 = \(\frac{12+12}{2}\) = 12;
T5 = \(\frac{15+12}{2}\) = 13\(\frac { 1 }{ 2 }\);
T6 = \(\frac{18+12}{2}\) = 15
Thus the required first six terms of A.P are;
7\(\frac { 1 }{ 2 }\), 9, 10\(\frac { 1 }{ 2 }\), 12, 13\(\frac { 1 }{ 2 }\) and 15
(iv) Given a = x and d = 3x + 2
We know that, Tn = a + (n – 1) d
⇒ Tn = x + (n – 1)(3x + 2)
∴ T1 = a = x
T2 = x + 3x + 2 = 4x + 2
T3 = x + 2(3x + 2) = 7x + 4
T4 = x + 3(3x + 2) = 10x + 6
T5 = x + 4(3x + 2) = 13x + 8
and T5 = x + 5(3x + 2) = 16x + 10
Thus required first six terms of A.P are;
x, 4x + 2, 7x + 4, 10x + 6, 13x + 8 and 16x + 10
Question 2.
Write the 5th and 8th terms of an A.P. whose 10th term is 43 and the common difference is 4.
Solution:
Let a be the first term and d be the common difference of given A.P.
∴ d = 4. Also T10 = 43
⇒ a + (10 – 1) d = 43
[∵ Tn = a + (n – 1)d]
⇒ a + 9d = 43 ⇒ a + 36 = 43 ⇒ a = 7
∴ T5 = a + (5 – 1)d = a + 4d = 7 + 4 × 4 = 23
T8 = a + 7d =7 + 7 × 4 = 35
Question 3.
In each of the following find the terms required.
(a) The seventh term of 2, 7, 12, ………
(b) The fifth term of 21, 28, 35, ………
(c) The eighteenth term of 9, 5, 1, …..
Solution:
(a) Given sequence be 2, 7, 12,
Here T2 – T1 = 7 – 2 = 5;
T3 – T2 = 12 – 7 = 5,
∴ T2 – T2 = T3 – T2 = …. = 5
Thus given sequence forms A.P. with first term a = 2 and common difference d= 5
We know that Tn = a + (n – 1) d
∴ T7 = a + (7 – 1) d = a + 6d = 2 + 6 × 5 = 32
(b) Given sequence be 21, 28, 35,….
Here, T2 – T1 = 28 – 21 = 7 ;
T3 – T2 = 35 – 28 = 7,….
∴ T2 – T1 =T3 – T2 = …. = 7
Thus, given sequence clearly forms an A.P. with first term a = 21 and common difference = d = 7
We know that Tn = a + (n – 1) d
∴ T5 = a + (5 – 1)d = a + 4d
= 21 + 4 × 7 = 49
(c) Given sequence be 9, 5, 1,….
Here, T2 – T1 = 5 – 9 = -4;
T3 – T2 = 1 – 5 = -4, ….
∴ T2 – T1 = T3 – T2 = …. = -4
Thus given sequence clearly forms an A.P with first term a = 9 and common diff. d = -4
We know that Tn = a + (n – 1)d
∴ T18 = 9 + (18 – 1)(- 4) = 9 – 68 = -59
Question 4.
Find the first four terms and the eleventh term of the series whose nth term is
(a) 4n – 2,
(b) 6n + 5,
(c) 101 – 3n
Solution:
(a) Given Tn = 4n – 2
∴ T1 = 4 – 2 = 2; T2 = 8 – 2 = 6;
T3 = 12 – 2 = 10 and T3 = 16 – 2 = 14
Thus T11 = 4 × 11 – 2 = 42
(b) Given Tn = 6n + 5
∴ T1 = 6 + 5 = 11 ; T2 = 6 × 2 + 5 = 17;
T3 = 6 × 3 + 5 = 23
and T4 = 6 × 4 + 5 = 29
∴ Tn = 6 × 11 + 5 = 71
(c) Given Tn = 101 – 3n
∴ T1 = 101 – 3 = 98; T2 = 101 – 6 = 95
T3 = 101 – 9 = 92 ; T4 = 101 – 12 = 89
∴ T11 = 101 – 33 = 68
Question 5.
The 5th term of an A.P. is 11 and the 9th term is 7. Find the 16th term.
Solution:
Let a be the first term and d be the common difference of given A.P. respectively it is given that
T5 = 11 ⇒ a + 4d – 11 …(1)
[∵ Tn = a + (n – 1)d]
and T9 ⇒ a + 8d = 7 …(2)
eqn. (1) – eqn. (2) gives ;
4d = – 4 and d = – 1
∴ from (1) ; a – 4 = 11 ⇒ a = 15
Thus T16 = a + 15d = 15 + 15 (- 1) = 0
Question 6.
Which term of the series 5, 8, 11, ……. is 320 ?
Solution:
Here, T2 – T1 = 8 – 5 = 3;
T3 – T2 = 11 – 8 = 3 …….
Clearly the given series form an A.P with first term a = 5 and common difference d = 3.
Let Tn of given A.P be 320.
∴ Tn = 320 ⇒ a + (n – 1) d = 320
⇒ 5 + (n- 1) 3 = 320 ⇒ (n – 1) 3 = 315
⇒ n- 1 = 105 ⇒ n = 106
Hence 106th term of given A.P be 320.
Question 7.
The fourth term of an A.P. is ten times the first. Prove that the sixth term is four times as great as the second term.
Solution:
Let a be the first term and d be the common difference of given A.P.
It is given that T4 = 10a
⇒ a + 3d = 10a ⇒ 9a = 3d ⇒ d = 3a
∴ T6 = a + 5d = a + 5 × 3a = 16a
and T2 = a + d = a + 3a = 4a
Clearly T6 = 4T2
Question 8.
The fourth term of an A.P. is equal to 3 times the first term, and the seventh term exceeds twice the third term by 1. Find the first term and the common difference.
Solution:
Let a be the first term and d be the common difference of an A.P.
According to given condition,
T4 = 3T1 ⇒ a + 3d = 3a
⇒ 2a = 3d …(1)
and T7 = 2T3 + 1
⇒ a + 6d = 2 (a + 2d) + 1
[∵ Tn = a(n – 1)d]
⇒ \(\frac { 3d }{ 2 }\) + 6d = 2 \(\left[\frac{3 d}{2}+2 d\right]\) + 1 [using (1)]
⇒ \(\frac { 15d }{ 2 }\) = 7d + 1 ⇒ \(\frac { d }{ 2 }\) = 1 ⇒ d = 2
∴ from (1); 2a = 3 × 2 ⇒ a = 3
Hence the first term of given A.P be 3 and common difference be 2.
Question 9.
Which term of the progression 19, 18\(\frac { 1 }{ 5 }\), 17\(\frac { 2 }{ 5 }\), ….. is the first negative term ?
Solution:
Here T2 – T1 = 18\(\frac { 1 }{ 5 }\) – 19 = –\(\frac { 4 }{ 5 }\)
and T3 – T2 = 17\(\frac { 2 }{ 5 }\) – 18\(\frac { 1 }{ 5 }\) = –\(\frac { 4 }{ 5 }\) and so on
Thus given progression forms an A.P. with first term a = 19
and common difference = d = –\(\frac { 4 }{ 5 }\)
Let Tn be the first negative term i.e. Tn < 0
⇒ a + (n – 1) d < 0
⇒ 19 + (n – 1) \(\left(-\frac{4}{5}\right)\) < 0
⇒ 95 – 4n + 4 < 0 ⇒ 99 < 4n ⇒ n > \(\frac { 99 }{ 4 }\) = 24\(\frac { 3 }{ 4 }\)
since n ∈ N
∴ n = 25
Hence 25th term of given A.P. be 24\(\frac { 3 }{ 4 }\).
Question 10.
(i) Find the value of k so that 8k + 4, 6k – 2, and 2k+ 1 will form an A.P.
(ii) Find a, b such that 7.2, a, b, 3 are in A.P. (SC)
Solution:
(i) Since 8k + 4,6k – 2 and 2k + 1 forms an A.P.
∴ 2 (6k – 2) = 8k + 4 + 2k + 7 [if a, b, c are in A.P Then 2b – a + c]
⇒ 12k – 4 = 10k + 1 ⇒ 2k = 15
⇒ k = \(\frac { 15 }{ 2 }\)
(ii) Since 7.2, a, b, 3 are in A.P
∴ 2a = 7.2 + b ⇒ 2a – b = 7.2 …(1)
From last three term of A.P. we have
2b = a + 3 ⇒ 2b – a = 3 …(2)
eqn. (1) + 2 eqn. (2); we have
2a – b + 4b – 2a = 7.2 + 6
⇒ 3b = 13.2 ⇒ b = 4.4
∴ from (1); 2a – 4.4 = 7.2 a = 5.8
Hence a = 5.8 and b = 4.4
Question 11.
Determine 2nd term and rth term of an A.P. whose 6th term is 12 and 8th term is 22.
Solution:
Let a be the first term and d be the common difference of an A.P respectively.
According to given conditions, we have
T6 = 12 ⇒ a + 5d = 12 …(1)
[∵ Tn = a + (n – 1) d]
and T8 = 22 ⇒ a + 7d =22 …(2)
eqn. (2) – eqn. (1) gives ;
2d = 10 ⇒ d = 5
∴ from (1) ; a + 25 = 12 ⇒ a = – 13
∴ T2 = a + d = – 13 + 5 = – 8
and Tr = a + (r – 1) d = – 13 + (r – 1) 5 = 5r – 18
Question 12.
Prove that the product of the 2nd and 3rd terms of an A.P. exceeds the product of the 1st and 4th by twice the square of the difference between the Inst and 2nd.
Solution:
Let a be the first term and d be the common difference of given A.P respectively.
We know that Tn = a + (n – 1) d
∴ T2 = a + d and T3 = a + 2d; T4 = a + 3d
Thus, T2 . T3 = (a + d) (a + 2d) = a2 + 3ad + 2d2
⇒ T2 . T3 = a (a + 3d) + 2d2
= T1T4 + 2(T2 -T1)2
Hence, the product of the 2nd and 3rd terms of an A.P exceeds the product of first and 4th term by twice the square of the difference between the 1st and 2nd term.
Question 13.
The 2nd, 31st and last term of an A.P. are 7\(\frac { 3 }{ 4 }\), \(\frac { 1 }{ 2 }\) and -6\(\frac { 1 }{ 2 }\) respectively. Find the first term and the number of terms.
Solution:
Let a be the first term and d be the common difference of given A.P.
Given T2 = 7\(\frac { 3 }{ 4 }\) ⇒ a + d = 7\(\frac { 3 }{ 4 }\) =\(\frac { 31 }{ 4 }\) …(1)
Also T31 = \(\frac { 1 }{ 2 }\) ⇒ a + 30d = \(\frac { 1 }{ 2 }\) …(2)
eqn. (2) – eqn. (1); we have
29d = \(\frac { 1 }{ 2 }\) – \(\frac { 31 }{ 4 }\) = –\(\frac { 29 }{ 4 }\) ⇒ d = –\(\frac { 1 }{ 4 }\)
∴ from (1); a = \(\frac { 31 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = 8
Let n be the required no. of terms of given
A.P. s.t. Tn = l = – 6\(\frac { 1 }{ 2 }\)
⇒ a + (n – 1) d = –\(\frac { 13 }{ 2 }\)
⇒ 8 + (n – 1) \(\left(-\frac{1}{4}\right)\) = –\(\frac { 13 }{ 2 }\)
⇒ (n – 1) \(\left(-\frac{1}{4}\right)\) = –\(\frac { 13 }{ 2 }\) – 8
⇒ (n – 1) \(\left(-\frac{1}{4}\right)\) = –\(\frac { 29 }{ 2 }\)
⇒ (n – 1) = –\(\frac { 29 }{ 2 }\) × (-4) = 58 ⇒ n = 59
Hence the required no. of terms of given A.P be 59.
Question 14.
If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, show that the 18th term of the A.P. is zero.
Solution:
Let a be the first term and d be the common difference of an A.P.
it is given that 7T7 = 11T11 …(1)
We know that Tn = a + (n – 1)d
∴ from (1) ; 7 (a + 6d) = 11 (a + 10d)
⇒ 4a + 68d = 0 ⇒4(a + 17d) = 0
⇒ a + 17d = 0 ⇒ T18 = 0
Question 15.
Determine k so that k + 2, 4k – 6 and 3k – 2 are three consecutive terms of an A.P.
Solution:
Given k + 2, 4k – 6 and 3k -2 are three consecutive terms of an A.P.
∴ 2(4k – 6) = k + 2 + 3k – 2
⇒ 8k – 12 = 4k ⇒ 4k = 12 ⇒ k = 3
Question 16.
The pth term of an A.P. is q and the fth term is p, show that the mth term is p + q – m.
Solution:
Let a be the first term and d be the common difference of given A.P respectively, it is given that Tp = q
⇒ a + (p – 1)d = q …(1)
and Tp = p ⇒ a + (q – 1) d = p …(2)
eqn. (2) – eqn. (1) gives ;
(q – p)d = p – q ⇒d = – 1
∴ from(1); a + (p – 1) (-1) = q
∴ a = p + q – 1
∴ Tm = a + (m – 1)d
= p + q – 1 + (m – 1) (-1)
= P + q – m
Question 17.
Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, Tm –\(\frac { 1 }{ n }\), Tn – \(\frac { 1 }{ m }\), then (a – d) equals
(a) \(\frac { 1 }{ mn }\)
(b) 1
(c) 0
(d) \(\frac { 1 }{ m }\) + \(\frac { 1 }{ n }\)
Solution:
Given a be the first term and d be the common difference of an A.P
given Tm = \(\frac { 1 }{ n }\) ⇒ a + (m – 1)d = \(\frac { 1 }{ n }\) …(1)
and Tn = \(\frac { 1 }{ m }\) ⇒ a + (n – 1)d = \(\frac { 1 }{ m }\) …(2)
eqn. (1) – eqn. (2) gives;
[m – 1 – n + 1] d = \(\frac { 1 }{ n }\) – \(\frac { 1 }{ m }\) = \(\frac { m – n }{ mn }\)
⇒ d = \(\frac { 1 }{ mn }\)
∴ from(1) ; a + \(\frac{(m-1)}{m n}\) = \(\frac { 1 }{ n }\)
⇒ a = \(\frac { 1 }{ n }\) – \(\frac{(m-1)}{m n}\)
⇒ a = \(\frac{m-m+1}{m n}\) = \(\frac { 1 }{ mn }\)
∴ a – d = \(\frac { 1 }{ mn }\) – \(\frac { 1 }{ mn }\) = 0
∴ Ans. (c)
Question 18.
Given that the (p + 1)th term of an A.P. is twice the (q + 1)th term, prove that the (3p + 1)th term is twice the (p + q + 1)th term.
Solution:
Let a be the first term and d be the common difference of an A.P.
given = Tp+1 = 2 Tq+1
⇒ a (p + 1 – 1) d = 2[a + (q + 1 – 1) d]
⇒ a + pd = 2 (a + qd)
⇒ a = (p – 2q)d …(1)
Now, T3p+1 = a + (3p + 1 – d)d
= a + 3pd
= (p – 2q) d + 3pd [using eqn. (1)]
= (p – 2q + 3p)d
= (4p – 2q) d
= 2(2p – q)d …(2)
Also, Tp+q+1 = a + (p + q + 1 – 1)d
= (p – 2q)d+(p + q)d
= (2P – q)d …(3)
From eqn. (2) and eqn. (3) ; we have
T3p+1 = 2 Tp+q+1