Sequences and Series Class 11 MCQ Online Test With Answers Questions

Answer: (a) AP
Hint:
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP

Answer: (b) 2b = a + c
Hint:
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Answer: (a) 2 + √3
Hint:
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}

Answer: (a) n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Answer: (b) (e – 1)² /2 e
Hint:
We know that,
e^x = 1 + x/1! + x² /2! + x³ /3! + x⁴ /4! + ………..
Now,
e¹ = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..
e^-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..
e¹ + e^-1 = 2(1 + 1/2! + 1/4! + ………..)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..
⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..
⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..
⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Answer: (b) 4⁵
Hint:
here it is given that T₃ = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar⁴
= a⁵r¹⁰
= (ar²)⁵
= 4⁵

Answer: (c) 1
Hint:
Let first term is a and the common difference is d of the AP
Now, T_m = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and T_n = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, T_mn = 1/mn + (mn-1)/mn
⇒ T_mn = 1/mn + 1 – 1/mn
⇒ T_mn = 1

Answer: (c) 6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b
Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Answer: (c) H.P.
Hint:
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² – 2pq}/(pq)²
⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}
⇒ -bc²/a³ = {b² – 2ca}/a²
⇒ -bc²/a = b² – 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c – 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Answer: (d) 350
Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a₁ = 1/2 + 1 = 3/2
a₂ = 2/2 + 1 = 2
a₃ = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a₃₅ = a₁ + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350

Answer: (c) 6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b
Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Answer: (c) 3
Hint:
Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar⁴
Now
⇒ ar² + ar⁴ = 90
⇒ a(r² + r⁴) = 90
⇒ r² + r⁴ = 90
⇒ r² ×(r² + 1) = 90
⇒ r²(r² + 1) = 3² ×(3² + 1)
⇒ r = 3
So the common ratio is 3

Answer: (b) 3775
Hint:
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)×{2a + (n – 1)d}
= (50/2)×{2×2 + (50 – 1)3}
= 25×{4 + 49×3}
= 25×(4 + 147)
= 25 × 151
= 3775

Answer: (b) 31/48
Hint:
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48

Answer: (a) n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Answer: (c) 740
Hint:
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3×7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2×4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2)×{2×(-1) + (20-1)×4}
= 10×{-2 + 19×4)}
= 10×{-2 + 76)}
= 10 × 74
= 740

Answer: (c) 11
Hint:
Given,
the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, ….
⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}
⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n – n = 56 – 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11

Answer: (b) 2abc
Hint:
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ………… 1
Again, given G₁ and G₁ are two GM between b and c,
⇒ b, G₁, G₂, c are in the GP having common ration r, then
⇒ r = (c/b)^1/(2+1) = (c/b)^1/3
Now,
G₁ = br = b×(c/b)^1/3
and G₁ = br = b×(c/b)^2/3
Now,
(G₁)³ + (G₂)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G₁)³ + (G₂)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G₁)³ + (G₂)³ = b³ ×(c/b)×( b + c)/b
⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b
⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b ………….. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G₁)³ + (G₂)³ = b² × c × (2a/b)
⇒ (G₁)³ + (G₂)³ = b × c × 2a
⇒ (G₁)³ + (G₂)³ = 2abc