Tips and Tricks and Shortcuts for Geometric Progression

December 13, 2022

Tips and Tricks for Geometric Progression

Geometric Progression is used in mathematics or aptitude, and they have various applications in physics, engineering, biology, economics, computer science, queueing theory, finance and geometric. Because in this sequence each successive term can be obtained by multiplying the previous term.That’s why Tips  and Tricks and Shortcuts of Geometric Progression is very important to know solve questions Easily.

Tips and Tricks and Shortcuts of Geometric Progression is given in this Page.In this page also different Types of Question is given that will help in Different Examination.

Tips and Tricks for Geometric Progression

Tips and Tricks on Geometric Progression easily

  • Definition 

In the Geometric Progression , the ratio of any term to its preceding term is constant in a sequence of numbers . Geometric Progression is a sequence of numbers where each term is calculated by multiplying the previous one by a fixed, non-zero number called the common ratio.

  • Here, are easy tips and tricks for you on Geometric Progression problems easily and efficiently in competitive exams.
  • There are 4 types of questions asked in exams

 

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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