How To Solve Geometric Progression Questions Quickly

How to Solve Geometric Progression Quickly​

In this Page you will learn How to solve Geometric Progression Quickly with different types of problems and their solutions.

How to Solve GP Questions Quickly

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.

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

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