How To Solve Geometric Progression Questions Quickly

How to solve geometric progression Quickly

How to Solve Geometric Progression Questions 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. For 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 , then 6 inches and  3 inches and so on , until the ball stops bouncing.

Definition & Formulas of GP

  • 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 Find Quickly 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 = arn-1

a3 = 45 * 0.23-1

a3 = 45 * 0.22

a3 = 45 * 0.04

a3 = 1.8

Correct option: C

 

Question 2. Find the nth term of the series -1/2 , 1/4 , -1/8……….. (Note: n is an even number)

Options:

A.    (1/2)n

B. 1/2 

C. 2n

 D.  n

Solution:     In the given series,

a = – 1/2

r = – 1/2       {1/4 / (-1/2)}

nth term = an = arn-1

an = (-1/2)*(-1/2)n-1

an= (-1/2 )n

However in the question it is given that n is even

Therefore,

an = (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  (a/r),  a,  ar

Sum of the three numbers =   (a/r) + a + ar = 14

Product of the three numbers =  (a/r) * a * ar = 64  i.e.  a3 =64

Therefore, a = 4

Put the value of a in  a/r + a + ar = 14

a ( 1/r + 1 + r) = 14

4( 1/r + 1 + r) = 14

2(1 + r + r2) = 7r

2r2 – 5r + 2 = 0

r = 2  or  1/2

If r = 2, then numbers are 2, 4, 8.
If r = 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 = 3/1 = 3,

an =729

an = arn-1

729 = 1* (3n-1)

729 = 3n-1

36 = 3n-1

6 = n-1

n = 7

Correct option: D

 

Question 2.  Find the last term of the series where, a = 3, r = 5, and number of terms are 5

Options:

A. 1875

B.625

C.1565

D.1785

Solution:    According to the question,

a = 3,

r = 5,

n = 5

an = arn-1

a5  = 3* (5 5-1)

a5= 3 * 54

a5= 3 * 625

a5= 1875

Correct option: A

 

Question 3. 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 * 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: Sn= [a1(rn-1)]/(r-1) and if r < 1 then Sn=[a1(1rn)]/(1r)

 

 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 = 8/16 = 0.5

a = 16

Sum of the terms for an infinite GP = a / (1 – r)

Thus, sum of the terms = 16/ (1 – 0.5) = 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

Sn = [a1(rn-1)] / (r-1)

S7 = [4(37-1)] / (3-1)

S7 = 4(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, Sn = {a1[(rn)-1]}/(r-1)

S6 =  {2[(46)-1]}/(4-1) = 2730

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)1/2

 

Question 1 Find the Geometric Mean (GM) between 4/6 and 169/6

Options:

A. 26/3

B.26/5

C. 26/6

D. 26/9

Solution:    We know that

GM = (ab)1/2

Therefore, GM =(4/6*169/6 )1/2 = (676/36)1/2 = 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 =  (a + b)/2

GM = (ab)1/2

Let the number be x and y

AM = (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 * 441

(x-y)2  = 22500 -1764

x-y   =   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)(1/3)

Therefore, GM =(10*15*20)(1/3)

GM = (3000)(1/3)

GM = 14.4 ~ 15

Correct option: C

 

Read Also – Formulas and Concept Of Geometric Progression