How To Solve Geometric Progression Questions Quickly

How to Solve Geometric Progression Quickly​

A Geometric Progression (GP), also known as Geometric sequence, is a sequence of non – zero numbers which differ from each other by a common ratio.We will learn How to Solve Geometric Progression Quickly by taking an example , a bouncing ball is an example of a finite geometric sequence. Each time the ball bounces it’s height gets cut down by half. If the ball’s first height is 4 feet, the next time it bounces it’s highest bounce will be at 2 feet, and then 1 feet , and so on , until the ball stops bouncing.

In this Page you will learn How to solve Geometric Progression Quickly with different Types of Problems and their Solutions. Definition of Geometric Progression

• A geometric progression (GP) is a sequence in which element after the first term is obtained by multiplying the preceding element by a constant called common ratio which is denoted by ‘r.’
• A GP is represented in the form a, ar, ar2, ar3, …….
where

a = the first term

r = the common ratio.

Type 1: How To Solve Geometric Progression Quickly for nth term of a GP i.e.  an = arn-1

Question 1. Find the 3th term of a GP if a= 45 and the common ration r = 0.2

Options:

A. 18

B. 8

C. 1.8

D. 20

Solution:     an = a1rn-1

a3 = 45 x 0.23-1

a3 = 45 x 0.22

a3 = 45 x 0.04

a3 = 1.8

Correct option: C

Question 2. Find the nth term of the series $\mathbf{\frac{-1}{2} , \frac{1}{4} , \frac{-1}{8}}$……….. (Note: n is an even number)

Options:

A. $\mathbf{(\frac{1}{2})^{n}}$

B.$\mathbf{(\frac{1}{2})}$

C. 2n

D.  n

Solution:     In the given series,

a = $\frac{- 1}{2}$

r=$\frac{-1}{2}(\frac{\frac{1}{4}}{\frac{-1}{2}})$

nth term = $a_{n} = ar^{n-1}$

$a_{n} = (\frac{-1}{2})\times (\frac{-1}{2})^{n-1}$

$a_{n} = (\frac{-1}{2})^{n}$

However in the question it is given that n is even

Therefore,

$a_{n} = (\frac{1}{2})^{n}$

Correct option: A

Question 3.The sum of three numbers is 14 and their product is 64. All the three numbers are in GP. Find all the three numbers when value of r is a whole number?

Options:

A. 3, 6, 4

B.2, 4, 8

C.2, 4, 6

D. 4, 4, 4

Solution:    The three numbers can be written as  $(\frac{a}{r}), a, ar$

Sum of the three numbers =   $(\frac{a}{r}) + a + ar = 14$

Product of the three numbers = $(\frac{a}{r}) \times a \times ar = 64$

i.e.  a3 =64

Therefore, a = 4

Put the value of a in $\frac{a}{r} + a + ar = 14$

$a (\frac{ 1}{r} + 1 + r) = 14$

$4(\frac{ 1}{r} + 1 + r) = 14$

2(1 + r + r2) = 7r

2r2 – 5r + 2 = 0

r = 2  or $\frac{1}{2}$

If r = 2, then numbers are 2, 4, 8.
If r = $\frac{1}{2}$ then numbers are 8, 4, 2

Correct option: B

Type 2: How To Solve Quickly GP Number of terms in the series: an = arn-1

Question 1. Find the number of terms in the series 1, 3, 9 , 27,……..729

Options:

A. 6

B. 8

C.5

D.7

Solution:    In the given series,

a1 = 1,

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

an =729

an = arn-1

729 = 1 x (3n-1)

729 = 3n-1

36 = 3n-1

6 = n-1

n = 7

Correct option: D

Question 2. Find the number of terms in the series where a1 = 3, a2 = 6, a3 = 12 an = 384

Options:

A. 8

B.9

C. 10

D. 7

Solution:     In the given series,

a1 = 3,

r= 2,

an = 384

an = arn-1

384 = 3 x 2n-1

128 = 2n-1

27 = 2n-1

n = 8

Correct option: A

Type 3: Quickly Solve Sum of first ‘n’ terms of the series of GP: if r>1 then $s_{n} = a \times \frac{r^{n} -1}{r-1}$  and if r < 1 then $s_{n} = a \times \frac{1-r^{n}}{1-r}$

Question 1. Find the sum up to infinity for the series 16, 8, 4, 2…

Options:

A. 23

B. 32

C. 64

D. 46

Solution:    In the given series

r=$\frac{8}{16} = 0.5$

a = 16

Sum of the terms for an infinite GP = $\frac{a}{ (1 – r)}$

Thus, sum of the terms = $\frac{16}{(1 – 0.5)} = \frac{16}{0.5} = 32$

Correct option: B

Question 2.  Find the sum of the series 4 + 12 + 36 + … up to 7 terms

Options:

A. 4372

B.1456

C.1500

D.3466

Solution:    In the given series a = 4,

r = 3,

n = 7

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

$s_{7} = 4 \times \frac{3^{7} -1}{3-1}$

$s_{7} = 4\times \frac{2187-1}{3-1}$

s7 = 4372

Correct option: A

Question 3.  Find the sum of the GP: 2, 8, 32, 128… which contains 6 terms in the series

Options:

A. 2370

B. 2073

C.2730

D. 2037

Solution:    Here a = 2,

r = 4 and

n = 6

We know that, $s_{n} = a \times \frac{r^{n} -1}{r-1}$

$s_{6} = 2 \times \frac{4^{6} -1}{4-1}$

Hence, the sum of the 6 series (2, 8, 32, 128… ) = 2730

Correct option: C

Type 4: How to Solve Quickly Geometric Mean (GM) of the series: GM =$(ab)^{\frac{1}{2}}$

Question 1 Find the Geometric Mean (GM) between $\frac{4}{6}$ and $\frac{169}{6}$

Options:

A. $\frac{26}{3}$

B. $\frac{26}{5}$

C. $\frac{26}{6}$

D. $\frac{26}{9}$

Solution:    We know that

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

Therefore, GM =$(\frac{676}{36})^{\frac{1}{2}}$

=$\frac{ 26}{6}$

Correct option: C

Question 2. The AM of two numbers is 75 and GM of two number is 21. Find the numbers.

Options:

A. 150 and 2

B.147 and 4

C. 145 and 3

D. 147 and 3

Solution:    We know that,

AM$\frac{(a + b)}{2}$

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

Let the number be x and y

AM = $\frac{(x+y)}{2} = 75$

x+y = 150……..(1)

GM = 21

Hence , ab = 441

We know that (x-y)2 = (x+y)2 – 4xy
(x-y)2  = (150)2 – 4 x 441

(x-y)2  = 22500 -1764

(x-y)2  =   20736

x – y = 144…..(2)

Adding both the equation (1) and (2)

x + y = 150

x – y = 144

x = 147

Putting the value of x in equation (2)

x – y = 144

147 – y = 144

y = 3

Correct option: D

Question 3. Find the approximate Geometric Mean (GM) 10, 15, 20

Options:

A. 51

B. 50

C. 15

D. 49

Solution:    We know that

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

Therefore, GM =$(10\times 15\times 20)^{\frac{1}{3}}$

GM = $(3000)^{\frac{1}{3}}$

GM = 14.4 ~ 15

Correct option: C