Multiple Choice Questions on Sequence and Series Class 11 OP Malhotra Exe-14K ISC Maths Solutions Ch-14. In this article you would learn to solve hard mcq’s questions easily on 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.

Sequence and Series Class 11 OP Malhotra Multiple Choice Questions ISC Math Solutions Ch-14

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-14	Sequence and Series
Writer	OP Malhotra
Exe-14(K)	Multiple Choice Questions.

Multiple Choice Questions on Sequence and Series

OP Malhotra ISC Class 11 Maths Solutions

Que-1: If T_n = n(n − 4)/(n² + 1), then the 9th term is:

(a) 55/82   (b) 40/82   (c) 45/82   (d) 36/82

Answer: (c) 45/82

Sol: Given T_n = n(n−4)/(n²+1). To find T₉, substitute n = 9.
So, T₉ = 9(9−4)/(9²+1) = 9×5/(81+1) = 45/82.

Que-2: The 30th term of A.P. 10, 7, 4, … is:

(a) −90   (b) −77   (c) −87   (d) −67

Answer: (b) −77

Sol: First term a = 10 and common difference d = 7−10 = −3.
Using formula Tₙ = a + (n−1)d,
T₃₀ = 10 + 29(−3) = 10 − 87 = −77.

Que-3: 3 + 5 + 7 + … to n terms is:

(a) n²   (b) n(n−2)   (c) n(n+2)   (d) (x+1)²

Answer: (c) n(n+2)

Sol: This is an A.P. where a = 3 and d = 2.
Sum of n terms: Sₙ = n/2 [2a + (n−1)d]
= n/2 [6 + 2(n−1)] = n/2 [2n + 4] = n(n+2).

Que-4: If T₂₅ = 15 and T₁₅ = 25, then T₄₀ is:

(a) −1   (b) −10   (c) −5   (d) 0

Answer: (d) 0

Sol: Using Tₙ = a + (n−1)d,
T₂₅ = a + 24d = 15 …(1)
T₁₅ = a + 14d = 25 …(2)
Subtract (1)-(2): 10d = −10 ⇒ d = −1.
Put in (2): a + 14(−1) = 25 ⇒ a = 39.
Now T₄₀ = a + 39d = 39 − 39 = 0.

Que-5: The 6th term of sequence 3, 1, 1/3,… is:

(a) 1/27   (b) 1/9   (c) 1/81   (d) 1/17

Answer: (c) 1/81

Sol: It is a G.P. with first term a = 3 and ratio r = 1/3.
Formula: Tₙ = a rⁿ⁻¹
T₆ = 3 × (1/3)⁵ = 3 / 243 = 1/81.

Que-6: If 100T₁₀₀ = 50T₅₀, then T₁₅₀ is: 

(a) 150   (b) 0   (c) −150   (d) 150 times T₅₀

Answer: (b) 0

Sol: Given: 100T₁₀₀ = 50T₅₀
⇒ 2T₁₀₀ = T₅₀
Using standard relation of sequence:
T₁₅₀ = T₁₀₀ + T₅₀ − 2T₁₀₀
Substitute: T₅₀ = 2T₁₀₀
⇒ T₁₅₀ = T₁₀₀ + 2T₁₀₀ − 2T₁₀₀ = 0

Que-7: In a G.P., each term equals sum of next two terms. Ratio is:

(a) √5   (b) (√5−1)/2   (c) √5/2   (d) (√5+1)/2 

Answer: (b) (√5−1)/2

Sol: Let terms be a, ar, ar²…
Given: a = ar + ar² ⇒ divide by a ⇒ 1 = r + r²
⇒ r² + r −1 = 0
⇒ r = (−1 ± √5)/2. Positive value ⇒ (√5−1)/2.

Que-8: Digits after decimal in 12th term of 32,16,8,…:

(a) 5   (b) 6   (c) 7   (d) 8

Answer: (b) 6

Sol: Sequence is G.P. with a = 32, r = 1/2.
T₁₂ = 32 × (1/2)¹¹ = 32/2048 = 1/64 = 0.015625.
It has 6 digits after decimal.

Que-9: If S₄ = −34, S₅ = −60, S₆ = −93, then (d, a) is:  

(a) −7,2   (b) 7,−4   (c) 7,−2   (d) 4,−7

Answer: (a) −7, 2

Sol: T₅ = S₅ − S₄ = −60 − (−34) = −26
T₆ = S₆ − S₅ = −93 − (−60) = −33
So d = T₆ − T₅ = −7
Now T₅ = a + 4d
⇒ −26 = a − 28
⇒ a = 2.

Que-10: If a = 148, d = −2 and A.M. = 125, then n is: 

(a) 18   (b) 24   (c) 30   (d) 36

Answer: (b) 24

Sol: A.M. = (first term + last term)/2
125 = (148 + l)/2
⇒ l = 102
Now l = a + (n−1)d
⇒ 102 = 148 − 2(n−1)
⇒ 2(n−1) = 46
⇒ n = 24.

Que-11: If A.M. = 3 and G.M. = 1, then x²+y² is :

(a) 30   (b) 31   (c) 32   (d) 34

Answer: (d) 34

Sol: A.M. = (x+y)/2 = 3 ⇒ x+y = 6
G.M. = √(xy) = 1
⇒ xy = 1
Now x² + y² = (x+y)² − 2xy
= 36 − 2 = 34.

Que-12: Product of 5 consecutive G.P. terms = 243/32. Middle term is:

(a) 2/3   (b) 3/2   (c) 4/3   (d) 3/4

Answer: (b) 3/2

Sol: Let middle term = m.
Then terms: m/r², m/r, m, mr, mr²
Product = m⁵ = 243/32
⇒ m = (243/32)^(1/5) = 3/2.

Que-13: If |x| < 1, then the sum of the series 1 + 2x + 3x² + 4x³ + …. ∞ will be:   

(a) 1/(1−x)   (b) 1/(1+x)   (c) 1/(1+x²)   (d) 1/(1−x)²

Answer: (d) 1/(1−x)²

Sol: We know that the sum of the infinite series
1 + x + x² + x³ + … = 1/(1−x), where |x| < 1.
Differentiate both sides with respect to x:
d/dx (1 + x + x² + x³ + …) = d/dx [1/(1−x)]
0 + 1 + 2x + 3x² + 4x³ + … = 1/(1−x)²
Hence, required sum = 1/(1−x)².

Que-14: If S_n = 1³ + 2³ + … + n³ and T_n = 1 + 2 + … + n, then: 

(a) S_n = T_n   (b) S_n = T_n⁴   (c) S_n = T_n²   (d) S_n = T_n

Answer: (c) S_n = T_n²

Sol: We know the formulas:
Sum of first n natural numbers: T_n = n(n+1)/2
Sum of cubes: S_n = [n(n+1)/2]²
Thus, S_n = (T_n)².

Que-15: The difference between two numbers is 48 and the difference between their arithmetic mean and geometric mean is 18. Then, the greater number is:   

(a) 96   (b) 60   (c) 54   (d) 49

Answer: (d) 49

Sol: Let numbers be a and b.
a − b = 48 ⇒ a = b + 48
AM = (a + b)/2,   GM = √(ab)
Given: AM − GM = 18
⇒ (b + 48 + b)/2 − √(b(b + 48)) = 18
⇒ (2b + 48)/2 − √(b² + 48b) = 18
⇒ b + 24 − √(b² + 48b) = 18
⇒ √(b² + 48b) = b + 6
Squaring both sides:
b² + 48b = (b + 6)² = b² + 12b + 36
⇒ 36b = 36 ⇒ b = 1
So, a = b + 48 = 49

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

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