# Geometric Progression Shortcut, tricks, and tips

## Geometric Progression tricks, shortcuts, and tips

Geometric Progression Tips,Tricks and Shortcuts are very important to know solve questions easily. In this page different types of question arevgiven that will help you in different examination. Here, are some easy tips and tricks for you to solve Geometric Progression questions quickly, easily , and efficiently in competitive exams.

### Geometric Progression Tips and Tricks and Shortcuts

• Three non-zero terms a, b, c are in GP if and only if b^2 = ac.
• In a GP, three consecutive terms can be taken as a/r, a, ar….
• If a, b and c are three quantities in GP, then their reciprocals (b/a), (c/b), and (c/a) are also in GP.
• If we multiply or divide each term of the GP by a non-zero quantity, then resulting sequence remains a GP with the same common ratio.

### Type 1: Find nth term of the series

Question 1: Find 11th term in the series 2,4,8,16 …

Options:

A. 2042

B. 2200

C. 1024

D. 2048

Solution:    We know that,

an = arn-1

where

r(common ratio) = $\frac{4}{2} = 2$

a1= first term = 2

an-1= the term before the nth term,

n = number of terms

In the given series,

r (common ratio) = $\frac{4}{2} = 2$

Therefore, 11th term = a11

a11 = 2 x 211-1

a11 = 2 x 210

a11 = 2 x 1024

a11 = 2048

Correct option: D

Question 2 : Find last term in the series if there are 7 term in this series 3,15,75,375 …

Options:

A. 46875

B. 44875

C. 42875

D. 40875

Solution:      We know that,

an = arn-1

where

r(common ratio) = $\frac{15}{3} = 5$

a1= first term = 3

an-1= the term before the nth term,

n = number of terms

In the given series,

r (common ratio) = $\frac{15}{3} = 5$

Therefore, 7th term = a7

a7 = 3 x 57-1

a7 = 46875

Correct option: A

### Type 2: Find number of terms in the series

Question 1. Find the number of terms in the GP 6, 12, 24, 48……1536

Options:

A. 6

B. 7

C. 9

D. 8

Solution:

We know that,

In the given series,

a1 = 6,

a2 = 12,

r = $\frac{12}{6} = 2$,

an = 1536

an = arn-1

1536 = 6 x 2n-1 (divide both side by 6)

256 = 2n-1

28 = 2n-1

8 = n – 1

n = 9

Therefore, there are 9 terms in the series.

Correct option: C

Question 2. Find the number of terms in the GP where a1 = 10, a2 = 40 ,a3 = 160 ,an = 10240

Options:

A. 10

B. 7

C. 9

D. 6

Solution:    We know that,

In the given series,

a1 = 10,

a2 = 40,

r = $\frac{40}{10} = 4$,

an = 10240

an = arn-1

10240 = 10 x 4n-1 (divide both side by 10)

1024 = 4n-1

45 = 4n-1

5 = n – 1

n = 6

Therefore, there are 6 terms in the series.

Correct option: D

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

Question 1. How many terms of the series 1 + 3 + 9 +….sum to 121

Options:

A. 18

B. 19

C. 13

D. 5

Solution:    We know that,

$s_{n} = a \times \frac{r^{n} -1}{r-1}$ if r>1

In the given series,

a = 1,

$r = \frac{3}{1} = 3$,

Sn = 121

121 =$1\times \frac{3^{n} -1}{3-1}$

121 = $\frac{3^{n} -1}{3-1}$

242 = (3n – 1)

243 = 3n

35 = 3n

n = 5

Correct option: D

Question 2. Find Sum of given Geometric Series upto 9th term 7,14,28,56……

Options:

A. 3177

B. 3577

C. 1377

D. 5377

Solution:    We know that,

$s_{n} = a \times \frac{r^{n} -1}{r-1}$ if r>1

In the given series,

a = 7,

$r = \frac{14}{7} = 2$,

Sn = $7\times \frac{2^{9} -1}{2-1}$

= 3577

Correct option: B

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

Question 1. What is the geometric mean of 2, 3, and 6?

Options

A. 4.5

B. 6.5

C. 3.30

D. 6.4

Solution:    We know that,

GM = $(abc)^{\frac{1}{3}}$

Therefore, there Geometric Mean (GM) = $(2\times 3\times 6)^{\frac{1}{3}}$

= 3.30

Correct option: C

Question 2 . What is the geometric mean of 36 and 9?

Options

A. 24

B. 16

C. 18

D. 14

Solution:

We know that,

GM = $(ab)^{\frac{1}{2}}$

Therefore, there Geometric Mean (GM) = $(36\times 9)^{\frac{1}{2}}$

= 18

Correct option: C

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