Progressions - Aptitude MCQ Questions and Solutions with Explanations

31

For an A.P. if a₂₅ - a₂₀ = 45, then d equals to:

A) 9 B) -9 C) 18 D) 23

Discuss

Answer & Solution

Answer: Option A

Solution:

a_n = a + (n - 1) × d
⇒ a₂₅ = a + 24d
and a₂₀ = a + 19d
a₂₅ - a₂₀ = 45
⇒ a + 24d - a - 19d = 45
⇒ 5d = 45
⇒ d = 9

32

For A.P. T₁₈ - T₈ = ........ ?

A) d B) 10d C) 26d D) 2d

Discuss

Answer & Solution

Answer: Option B

Solution:

T_n = a + (n - 1) × d
T₁₈ = a + 17d
T₈ = a + 7d
T₁₈ - T₈ = 17d - 7d
= 10d

33

Which term of the A.P. 24, 21, 18, ............ is the first negative term?

A) 8^th B) 9^th C) 10^th D) 12^th

Discuss

Answer & Solution

Answer: Option C

Solution:

a_n = a + (n - 1) × d where d = -3, Let a_n = 0
⇒ 0 = 24 + (n - 1) × -3
⇒ 0 = 24 - 3n + 3
⇒ 3n = 27
⇒ n = 9
⇒ 10^th term will be negative (-ve)

34

15^th term of A.P., x - 7, x - 2, x + 3, ........ is

A) x + 63 B) x + 73 C) x + 83 D) x + 53

Discuss

Answer & Solution

Answer: Option A

Solution:

a_n = a + (n - 1) × d
where,
d = x - 2 -x + 7 = 5,
a = x - 7
⇒ a₁₅ = (x - 7) + (15 - 1) × 5
⇒ a₁₅ = (x - 7) + 14 × 5
⇒ a₁₅ = x - 7 + 70
⇒ a₁₅ = x + 63

35

If an A.P. has a = 1, t_n = 20 and s_n = 399, then value of n is :

A) 20 B) 32 C) 38 D) 40

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {S_n} = \frac{1}{2}\left( {a + l} \right) \times n \cr & \Rightarrow 399 = \left( {1 + 20} \right) \times \frac{n}{2} \cr & \Rightarrow 399 \times 2 = 21 \times n \cr & \Rightarrow n = 399 \times \frac{2}{{21}} \cr & \Rightarrow n = 19 \times 2 \cr & \Rightarrow n = 38 \cr} $$

36

The sum of first five multiples of 3 is:

A) 45 B) 65 C) 75 D) 90

Discuss

Answer & Solution

Answer: Option A

Solution:

$${S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2}$$
$$ \Rightarrow {S_5} = \left[ {2 \times 3 + \left( {5 - 1} \right)3} \right]$$ $$ \times \frac{5}{2}$$
$$\eqalign{ & \Rightarrow {S_5} = \left[ {6 + 12} \right] \times \frac{5}{2} \cr & \Rightarrow {S_5} = 18 \times \frac{5}{2} \cr & \Rightarrow {S_5} = 9 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45 \cr} $$

37

(1) + (1 + 1) + (1 + 1 + 1) + ....... + (1 + 1 + 1 + ...... n - 1 times) = ......

A) $$\frac{{n\left( {n + 1} \right)}}{2}$$ B) $$\frac{{n\left( {n - 1} \right)}}{2}$$ C) $${n^2}$$ D) $$n$$

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2} \cr & {\text{Here,}} \cr & d = 1 \cr & a = 1\, \cr & {\text{and}}\,\,n - 1\,\,{\text{terms}} \cr} $$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + \left( {n - 1 - 1} \right)} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + n - 2} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n\left( {n - 1} \right)}}{2}$$

38

In an A.P., if d = -4, n = 7, a_n = 4, then a is

A) 6 B) 7 C) 20 D) 28

Discuss

Answer & Solution

Answer: Option D

Solution:

a_n = a + (n - 1) × d
⇒ 4 = a + (7 - 1) × -4
⇒ 4 = a - 24
⇒ a = 4 + 24
= 28

39

Which term of the A.P. 92, 88, 84, 80, ...... is 0?

A) 23 B) 32 C) 22 D) 24

Discuss

Answer & Solution

Answer: Option D

Solution:

a_n = a + (n - 1) × d where d = -4, Let a_n = 0
⇒ 0 = 92 + (n - 1) × -4
⇒ 0 = 92 - 4n + 4
⇒ 4n = 96
⇒ n = 24

40

If a + 1, 2a + 1, 4a - 1 are in A.P., then the value of a is:

A) 1 B) 2 C) 3 D) 4

Discuss

Answer & Solution

Answer: Option B

Solution:

Let 1^st term = x = a + 1,
2^nd term = y = 2a + 1 and
3^rd term = z = 4a - 1
⇒ y - x = z - y
⇒ 2y = x + z
⇒ 2(2a + 1) = a + 1 + 4a - 1
⇒ 4a + 2 = 5a
⇒ a = 2