MCQ: Arithmetic Progressions - 2 - SSC CGL MCQ

a = 10, d = 10
a(1) = a = 10
a(2) = a1 + d = 10 + 10 = 20
a(3) = a2 + d = 20 + 10 = 30
a(4) = a3 + d = 30 + 10 = 40

a² = 13 and
a(4) = 3
The nth term of an AP;
a(n) = a + (n − 1) x d
a(2) = a + (2 - 1) x d
13 = a + d ………………. (i)
a(4) = a + (4 - 1) x d
3 = a + 3d ………….. (ii)
Subtracting equation (i) from (ii), we get
– 10 = 2 x d
d = – 5
put value of d in eq 1
13 = a + (-5)
a = 18
a(3) = 18 + (3 - 1) x (-5)
= 18 + 2 x (-5)
= 18 - 10 = 8

a(2) = 13
a + d = 13
a = 13 – d…. (i)
a(5) = 25
a + 4d = 25….(ii)
Substituting the value of (i) in (ii),
13 – d + 4 x d = 25
3 x d = 12
d = 4
So, a = 13 – 4 = 9
a(7) = a + 6d = 9 + 6(4) = 9 + 24 = 33

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

A.P. is 3, 8, 13, …, 253
d= 5.
Starting from the end of the series,
253, 248, 243, …, 13, 8, 5
a = 253
d = 248 − 253 = −5
n = 20
a(20) = a + (20 − 1) x d
a(20) = 253 + (19) x (−5)
a(20) = 253 − 95
a(20) = 158

Here a = 1 and d = 4
The n^th term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3

n^th term of the AP is 7n – 9
T_n = 7n – 1
T₁ = 7(1) – 1 = 6
The first term of AP is 6.

In this case since we have to find the 99^th term from the end.
We will consider the first term to be -1139 and the common difference will be 11
Now, a = -1139, d = 11 and n = 99
T₉₉ = a + (n – 1)d
T₉₉ = -1139 + (99 – 1)11
T₉₉ = -1139 + 1078 = -61
The value of 99^th term from the end is -61. 

5(5^th term) = 15(15^th term)
5^th term = a + (5 – 1)d = a + 4d
15^th term = a + (15 – 1)d = a + 14d
5(a + 4d) = 15(a + 14d)
5a + 20d = 15a + 210d
20d – 210d = 15a – 5a
-190d = 10a
a = -19d
Now, the 20^th term = a + (20 – 1)d = a + 19d
But a = -19d
Hence, -19d + 19d = 0 

Here 7^th term = 20
11^thterm = 40
Let the first term of the AP be a and common difference be d
T₇ = a + (n – 1)d = 20
T₇ = a + (7 – 1)d = 20
T₇ = a + 6d = 20     (1)
T₁₁ = a + (n – 1)d = 40
T₁₁ = a + (11 – 1)d = 40
T₁₁ = a + 10d = 40     (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5 

n^th term of the AP is 2n + 7
T_n = 2n + 7
Common difference = T₂ – T₁ = (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2 

n^th term of the AP is 8n + 1
T_n = 8n + 1
T₂₀ = 8(20) + 1
T₂₀ = 161 

p^th term = q
a + (p – 1)d = q
a + pd – d = q     (1)
q^th term = p
a + (q – 1)d = p
a + qd – d = p     (2)
Subtracting (2) from (1) we get,
a + qd – d – (a + pd – d) = p – q
qd – pd = p – q
d = -1
Substituting in equation 1, we get,
a = p + q – 1
(p + q)^th term = a + (n – 1)d = p + q – 1 + (p + q – 1)(-1) = p + q – 1 – p – q + 1 = 0 

Here a = 5, d = 8 – 5 = 3 and n = 14
T₁₄ = a + (n – 1)d
T₁₄ = 5 + (14 – 1)3
T₁₄ = 5 + 13 × 3
T₁₄ = 44 

Here 11^th term = 1/13
13^th term = 1/11
Let the first term of the AP be a and common difference be d

Subtracting (1) from (2)
We get,

Now, substituting value of d in equation 1
We get,