Tips and Tricks: Geometric Progressions | Quantitative Aptitude for SSC CGL PDF Download

Type 1: Find nth term of the series

Example 1: Find 11^th term in the series 2,4,8,16 …
(a) 2042
(b) 2200
(c) 1024
(d) 2048
Ans: (d)
We know that,
a_n = ar^n-1
where
r(common ratio) = 4/2 = 2
a₁= first term = 2
a_n-1 = the term before the n^th term,
n = number of terms
In the given series,
r (common ratio) = 4/2 = 2
Therefore, 11^th term = a₁₁ 
a₁₁ = 2 x 2^11-1
a₁₁ = 2 x 2¹⁰
a₁₁ = 2 x 1024
a₁₁ = 2048

Example 2: Find last term in the series if there are 7 term in this series 3,15,75,375 …
(a) 46875
(b) 44875
(c) 42875
(d) 40875
Ans: (a)
We know that,
a_n = ar^n-1
where
r(common ratio) = 15/3 = 5
a₁= first term = 3
a_n-1= the term before the n^th term,
n = number of terms
In the given series,
r (common ratio) = 15/3 = 5
Therefore, 7^th term = a₇
a₇ = 3 x 5^7-1
a₇ = 46875

Type 2: Find number of terms in the series

We know that,
In the given series,
a₁ = 10,
a₂ = 40,
r = 40/10 = 4
a_n = 10240
a_n = ar^n-1
10240 = 10 x 4^n-1 (divide both side by 10)
1024 = 4^n-1
4⁵ = 4^n-1
5 = n – 1
n = 6
Therefore, there are 6 terms in the series.

Type 3: Related to sum of first ‘n’ terms of the Geometric series

Example 1: How many terms of the series 1 + 3 + 9 +….sum to 121
(a) 18
(b)19
(c) 13
(d) 5
Ans: (d)
We know that,

In the given series,
a = 1,
r = 3/1 = 3,
S_n = 121

242 = (3ⁿ – 1)
243 = 3ⁿ
3⁵ = 3ⁿ
n = 5

Example 2: Find Sum of given Geometric Series upto 9th term 7,14,28,56……
(a) 3177
(b) 3577
(c) 1377
(d) 5377
Ans: (b)
We know that,

 In the given series,
 a = 7,
r = 14/7 = 2

= 3577

Type 4: Find the Geometric Mean (GM) of the series.

Example 1: What is the geometric mean of 2, 3, and 6?  
(a) 4.5
(b) 6.5
(c) 3.30
(d) 6.4
Ans: (c)
We know that,
GM = (abc)^1/3
Therefore, there Geometric Mean (GM) = (2 x 3 x 6)^1/3
= 3.30

Example 2: What is the geometric mean of 36 and 9?  
(a) 24
(b) 16
(c) 18
(d) 14
Ans: (c)
We know that,
GM = (ab)1/2
 Therefore, there Geometric Mean (GM) = (36 x 9)^1/2
= 18