OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Ex 14(a)

Students often turn to Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(a) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(a)

Question 1.
Write the first five terms of the sequence using the given rule. In each case, the initial value of the index is 1.
(i) a_n = 2n
(ii) a_n = 3n – 2
(iii) a_n = n² + 5;
(iv) a_n = \(\frac{(-1)^{n-1}}{n^3}\)
(v) a_n = nth prime number for all natural numbers n.
Solution:
(i) Given a_n = 2n
For n = 1 ; a₁ = 2 × 1 = 2;
For n = 2; a₂ = 2 × 2 = 4;
For n = 3; a₃ = 2 × 3 = 6;
For n = 4; a₄ = 2 × 4 = 8;
For n = 5; a₅ = 2 × 5 = 10;
Hence first five terms of the sequence are ; 2, 4, 6, 8 and 10.

(ii) Given a_n = 3n – 2
∴ a₁ = 3 × 1 – 2 = 1;
a₂ = 3 × 2 – 2 = 4;
a₃ = 3 × 3 – 2 = 7;
a₄ = 3 × 4 – 2 = 10;
a₅ = 3 × 5 – 2 = 13
Hence the required first five terms of the sequence are 1, 4, 7, 10, 13.

(iii) a_n = n² + 5
∴ a₁ = 1² + 5 = 6; a₂ = 2² + 5 = 9;
a₃ = 3² + 5 = 14; a₄ = 4² + 5 = 21
and a₅ = 5² + 5 = 30
Hence, the required first five terms of the sequence are 6, 9, 14, 21, 30.

(iv) Given a_n = \(\frac{(-1)^{n-1}}{n^3}\)
∴ a₁ = \(\frac{(-1)^0}{1^3}\) = 1;
a₂ = \(\frac{(-1)^1}{2^3}\) = – \(\frac { 1 }{ 2 }\);
a₃ = \(\frac{(-1)^2}{3^3}\) = \(\frac { 1 }{ 27 }\);
a₄ = \(\frac{(-1)^3}{4^3}\) = – \(\frac { 1 }{ 64 }\);
and a₅ = \(\frac{(-1)^4}{5^3}\) = – \(\frac { 1 }{ 125 }\);
Hence the required first five terms of sequence are
1, –\(\frac { 1 }{ 8 }\), \(\frac { 1 }{ 27 }\), –\(\frac { 1 }{ 64 }\) and \(\frac { 1 }{ 125 }\)

(v) Given a_n = nth prime number for all natural numbers n
∴ a₁ = Ist prime no. = 2
a₂ = 2nd prime no. = 3
₃ = 3rd prime no. = 5
a₄ = 4th prime no. = 7
and a₅ = 5th prime no. = 11
Hence required first five terms of sequence are ; 2, 3, 5, 7 and 11 .

Question 2.
Write the first four terms of the sequence whose nth term is given
(i) \(\frac{2 n+1}{2 n-1}\)
(ii) \(\frac{n^2+1}{n}\)
(iii) \(\frac{2^n}{n^2}\)
(iv) sinⁿ 30°;
(v) (-1)ⁿ sin\(\frac{n \pi}{2}\);
(vi) (-1)^n-1 cos\(\frac{n \pi}{4}\);
Solution:
(i) Given a_n = \(\frac{2 n+1}{2 n-1}\)
∴ a₁ = \(\frac{2+1}{2-1}\) = 3;
a₂ = \(\frac{4+1}{4-1}\) = \(\frac{5}{3}\)
a₃ = \(\frac{6+1}{6-1}\) = \(\frac{7}{5}\) and
a₄ = \(\frac{8+1}{8-1}\) = \(\frac{9}{7}\)
Thus first four terms of sequence are 3, \(\frac{5}{3}\), \(\frac{7}{5}\) and \(\frac{9}{7}\).

(ii) Given a_n = \(\frac{n^2+1}{n}\)
∴ a₁ = \(\frac{1^2+1}{1}\) = 2;
a₂ = \(\frac{2^2+1}{2}\) = \(\frac{5}{2}\)
a₃ = \(\frac{3^2+1}{3}\) = \(\frac{10}{3}\) and
a₄ = \(\frac{4^2+1}{4}\) = \(\frac{17}{4}\)
Hence , the first four terms of sequence are 2, \(\frac{5}{2}\), \(\frac{10}{3}\) and \(\frac{17}{4}\)

(iii) Given a_n = \(\frac{2^n}{n^2}\)
∴ a₁ = \(\frac{2^1}{1^2}\) = 2;
a₂ = \(\frac{2^2}{2^2}\) = 1;
a₃ = \(\frac{2^3}{3^2}\) = \(\frac{8}{9}\)
a₄ = \(\frac{2^4}{4^2}\) = 1;
Thus required first four terms of sequence are 2, 1, \(\frac{8}{9}\), 1.

(iv) Given a_n = sinⁿ 30°
∴ a₁ = sin 30° = \(\frac{1}{2}\);
a₂ = sin² 30° = (sin 30°)² = \(\left(\frac{1}{2}\right)^2\) = \(\frac{1}{4}\);
a₃ = sin³ 30° = \(\left(\frac{1}{2}\right)^3\) = \(\frac{1}{8}\);
and a₄ = sin⁴ 30° =\(\left(\frac{1}{2}\right)^4\) = \(\frac{1}{16}\);
Hence, the required first four terms of sequence are \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\).

(v) Given a_n = (-1)ⁿ sin\(\frac{n \pi}{2}\)
∴ a₁ = (-1)¹ sin\(\frac{\pi}{2}\) = -1;
a₂ = (-1)² sin π = 0;
a₃ = (-1)³ sin\(\frac{3 \pi}{2}\) = (-1) × (-1) = 1;
a₄ = (-1)⁴ sin 2π = 0
Thus, the required first four terms of sequence are – 1, 0, 1, 0.

(vi) Given a_n = (-1)^n-1 cos \(\frac{n \pi}{4}\)
∴ a₁ = (-1)° cos \(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\);
a₂ = (-1)^2-1 cos \(\frac{2 \pi}{4}\) = 0
a₃ = (-1)² cos \(\frac{3 \pi}{4}\) = cos \(\left(\pi-\frac{\pi}{4}\right)\)
= – cos \(\frac{\pi}{4}\) = – \(\frac{1}{\sqrt{2}}\)
and a₄ = (-1)³ cos\(\frac{4 \pi}{4}\) = (-1) cos π = (-1) (-1) = 1
Thus, required first four terms of sequence are \(\frac{1}{\sqrt{2}}\), 0, – \(\frac{1}{\sqrt{2}}\), 1.

Question 3.
Find the first 4 terms and the 20th term of the sequence whose S_n = \(\frac{3}{2}\) (3ⁿ – 1).
Solution:
Given S_n = \(\frac{3}{2}\) (3ⁿ – 1)
∴ S_n-1 = \(\frac{3}{2}\) [3^n-1 – 1]
Thus, T_n = S_n – S_n-1
= \(\frac{3}{2}\) [3n – 1 – 3^n-1 + 1]
⇒ T_n = \(\frac{3}{2}\) (3ⁿ – 3^n-1)
∴ T₁ = \(\frac{3}{2}\) (3¹ – 3⁰) = \(\frac{3}{2}\) (3 – 1) = 3
T₂ = \(\frac{3}{2}\)(3² – 3) = \(\frac{3}{2}\) × 6 = 9
T₃ = \(\frac{3}{2}\) (3³ – 3²) = \(\frac{3}{2}\) × 18 = 27
and T₄ = \(\frac{3}{2}\) (3⁴ – 3³) = \(\frac{3}{2}\) × (81 – 27)
= \(\frac{3}{2}\) × 54 = 81
and T₂₀ = \(\frac{3}{2}\) (3²⁰ – 3¹⁹)
= \(\frac{3}{2}\) × 3¹⁹ (3 – 1) = 3²⁰

Question 4.
Find the 10th term of the sequence whose sum to n terms is 6n² + 7.
Solution:
Given S_n = 6n² + 7
∴ S_n-1 = 6(n – 1)² + 7
∴ T_n = S_n – S_n-1 = (6n² + 7) – [6 (n – 1)² + 7]
⇒ T_n = 6[n² – n² + 2n – 1] = 6(2n – 1) …(1)
putting n = 10 in eqn. (1); we have
T₁₀ = 6 (2 × 10 – 1) = 6 × 19 = 114