MCQ: Arithmetic Progressions - 1 - SSC CGL MCQ

AP: 10, 6, 2, …
a = 10, d = - 4
Sum of first n terms = S(n) = (n/2) x [2a + (n – 1) x d]
S5 = (5/2) x [2 x (10) + (5 – 1) x (-4)]
= (5/2) x [20 + 4 x (-4)]
= (5/2) x (20 – 16)
= (5/2) x (4)
= 5 x 2
= 10

4’s multiples are:
12, 16, 20, 24, …
a = 12 and d = 4
2 is the remainder for 250 / 4.
250 – 2 = 248 which is divisible by 2.
12, 16…248
nth term, a(n) = 248
a(n) = a + (n − 1) x d
248 = 12 + (n-1) × 4
236 / 4 = n - 1
59 = n - 1
n = 60

Gauss is the great mathematician who is credited with discovering the total value of the first 100 natural numbers.

for a given AP, the Nth term is:
a(n) = a + (n-1) x d
a(17) = a + (17−1) x d
a(17) = a + 16 x d
a(10) = a+9(d)
a(17) – a(10) = 7
(a + 16 x d) − (a + 9 x d) = 7
7 x d = 7
d = 1

a(18) – a(14) = 32
a(n) = a + (n – 1) x d
a + (17) x d – (a + 13 x d) = 32
(17) x d – (13) x d = 32
(4) x d = 32
d = 8

 a = -3 and a(2) = 4
a = -3
d = 4 - a = 4 - (-3) = 7
a(21)=a + (21-1) x d
= -3 + (20) x 7
= -3 + 140
= 137

AP: 21, 42, 63, 84, …
a = 21
d = 42 – 21 = 21
a(n) = 210
a + (n – 1) x d = 210
21 + (n – 1) x (21) = 210
21 + 21 x n – 21 = 210
21 x n = 210
n = 10

Given, 3, 8, 13, 18, … is the AP.
a = 3
d = a(2) – a(1) = 8 − 3 = 5
Let a(n) term be 78.
a(n) = a + (n − 1) x d
78 = 3 + (n − 1) x 5
75 = (n − 1) x 5
(n − 1) = 15
n = 16

–12, –6, 0, 6,…
Let a(1) = -12, a(2) = -6, a(3) = 0, a(4) = 6
First relational d,
a(2) – a(1) = -6 – (-12) = 6
Second relational d,
a(3) – a(2) = 0 – (-6) = 6
Third relational d,
a(4) – a(3) = 6 – (0) = 6
all the d are equals to each other, hence
d = 6.

The A.P. is -3, -1/2, 2 …
Where a = – 3
d = a(2) – a(1) = (-1 / 2) - (-3)
⇒ (-1 / 2) + 3 = 5 / 2
a(n) = a + (n−1) x d
a(11) = 3 + (11-1) x (5 / 2)
a(11) = 3 + (10) x (5 / 2)
a(11) = -3 + 25
a(11) = 22

d = -4, n = 7, a(n) = 4
a(n) = a + (n – 1) x d
a + (7 – 1) x (-4) = 4
a + 6 x (-4) = 4
a – 24 = 4
a = 4 + 24
a = 28

A.P. given in the question is 10, 7, 4, …
a = 10
d = a(2) – a(1) = 7 − 10 = −3
a(n) = a + (n−1) x d
a(30) = 10 + (30−1) x (−3)
a(30) = 10 + (29) x (−3)
a(30) = 10 − 87 = −77

The given series : 5, 8, 11, 14, …
a = 5
d = 8 – 5 = 3
a(n) = a + (n – 1) x d
10th term = a(10) = a + (10 – 1) x d
= 5 + 9 x (3)
= 5 + 27
= 32

a = 3
d = a(2) – a(1)
⇒ 1 – 3 = -2
⇒ d = -2

The starting four multiples of 2 is 2, 4, 6, 8
a=2 and d=2
n=4
S(n) = (n / 2) x [2a + (n - 1) x d]
S(4) = (4 / 2) x [2 x (2) + (4 - 1) x 2]
= (2) x [4 + 6]
= (2) x [10]
= 20