Sequence and Series Class 11 OP Malhotra Exe-14A ISC Maths Solutions Ch-14 Solutions. In this article you would learn about General /nTh term of a Sequence and Series. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Sequence and Series Class 11 OP Malhotra Exe-14A ISC Maths Solutions Ch-14

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-14	Sequence and Series
Writer	O.P. Malhotra
Exe-14(A)	General nTh term of a Sequence and Series.

Exercise- 14A

Sequence and Series Class 11 OP Malhotra Exe-14A Solution.

Que-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) an = [(−1)^(n−1)]/n³
(v) a_n = nth prime number for all natural numbers n.

Sol: (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 an = [(−1)^(n−1)]/n³
∴ a1 = {(−1)^0}/1³ = 1;
a2 = {(−1)¹}/2³ = –1/2;
a3 = (−1)²/3³ = 1/27;
a4 = (−1)³/4³ = – 1/64;
and a5 = {(−1)^4}/5³ = – 1/125;
Hence the required first five terms of sequence are
1, –1/8, 1/27, –1/64 and 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 .

Que-2: Write the first four terms of the sequence whose nth term is given
(i) (2n+1)/(2n−1)
(ii) (n²+1)/n
(iii) (2^n)/n²
(iv) sinⁿ 30°;
(v) (-1)ⁿ sin (nπ/2)
(vi) (-1)^n-1 cos (nπ/4)

Sol: (i) Given an = (2n+1)/(2n−1)
∴ a1 = (2+1)/(2−1) = 3;
a2 = (4+1)/(4−1) = 5/3
a3 = (6+1)/(6−1) = 7/5 and
a4 = (8+1)/(8−1) = 9/7
Thus first four terms of sequence are 3, 5/3, 7/5 and 9/7.

(ii) Given an = (n²+1)/n
∴ a1 = (1²+1)/1 = 2;
a2 = (2²+1)/2 = 5/2
a3 = (3²+1)/3 = 10/3 and
a4 = (4²+1)/4 = 17/4
Hence , the first four terms of sequence are 2, 5/2, 10/3 and 17/4

(iii) Given an = (2^n)/n²
∴ a1 = (2¹)/1² = 2;
a2 = 2²/2² = 1;
a3 = 2³/3² = 8/9
a4 = (2^4)/4² = 1;
Thus required first four terms of sequence are 2, 1, 8/9, 1.

(iv) Given an = sin^n 30°
∴ a1 = sin 30° = 1/2;
a2 = sin² 30° = (sin 30°)² = (1/2)² = 14;
a3 = sin³ 30° = (1/2)³ = 1/8;
and a4 = sin^4 30° =(1/2)^4 = 1/16;
Hence, the required first four terms of sequence are 1/2, 1/4, 1/8, 1/16.

(v) Given an = (-1)^n sin(nπ/2)
∴ a1 = (-1)¹ sin(π/2) = -1;
a2 = (-1)² sin π = 0;
a3 = (-1)³ sin(3π/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 an = (-1)^(n-1) cos (nπ/4)
∴ a1 = (-1)° cos (π/4) = 1/√2;
a2 = (-1)^(2-1) cos (2π/4) = 0
a3 = (-1)² cos (3π/4) = cos (π−(π/4))
= – cos π/4 = – 1/√2
and a4 = (-1)³ cos (4π/4) = (-1) cos π = (-1) (-1) = 1
Thus, required first four terms of sequence are 1/√2, 0, – 1/√2, 1.

Que-3: Find the first 4 terms and the 20th term of the sequence whose S_n = (3/2)  (3ⁿ – 1).

Sol: Given Sn = (3/2) (3ⁿ – 1)
∴ S_n-1 = (3/2)  [3^n-1 – 1]
Thus, T_n = S_n – S_n-1
= (3/2) [3n – 1 – 3^n-1 + 1]
⇒ T_n = (3/2)  (3ⁿ – 3^n-1)
∴ T₁ = (3/2)  (3¹ – 3⁰) = (3/2)(3 – 1) = 3
T₂ = (3/2) 2(3² – 3) = (3/2) × 6 = 9
T₃ = (3/2) (3³ – 3²) = (3/2) × 18 = 27
and T₄ = (3/2) (3⁴ – 3³) = (3/2) × (81 – 27)
= (3/2) × 54 = 81
and T₂₀ = (3/2) (3²⁰ – 3¹⁹)
= (3/2) × 3¹⁹ (3 – 1) = 3²⁰

Que-4: Find the 10th term of the sequence whose sum to n terms is 6n² + 7.

Sol: 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

–: End of Sequence and Series Class 11 OP Malhotra Exe-14A ISC Maths Ch-14 Solutions. :–

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