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# How To Solve Geometric Progression Questions 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, ar**, …….^{2}, ar^{3}

where

**a** = the first term

**r** = the common ratio.

**Type 1: How To Find Quickly ****n**^{th }term of a GP i.e. a_{n} = ar^{n-1}

^{th }term of a GP i.e.

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

**Options:**

**A. 18**

**B. 8**

**C. 1.8**

**D. 20**

**Solution: **a_{n} = ar^{n-1}

a_{3} = 45 * 0.2^{3-1}

a_{3} = 45 * 0.2^{2}

a_{3} = 45 * 0.04

a_{3} = 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. 2 ^{n}**

** D. n**

**Solution: **In the given series,

a = – 1/2

r = – 1/2 {1/4 / (-1/2)}

n^{th} term = an = ar^{n-1}

an = (-1/2)*(-1/2)^{n-1}

a_{n}= (-1/2 )^{n}

However in the question it is given that n is even

Therefore,

a_{n} = (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. a^{3 }=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 + r^{2}) = 7r

2r^{2} – 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: **a_{n} = ar^{n-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,

a_{1} = 1,

r = 3/1 = 3,

a_{n }=729

a_{n} = ar^{n-1}

729 = 1* (3^{n-1})

729 = 3^{n-1}

3^{6} = 3^{n-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

a_{n} = ar^{n-1}

a_{5} = 3* (5 ^{5-1})

a_{5}= 3 * 5^{4}

a_{5}= 3 * 625

a_{5}= 1875

**Correct option: A**

**Question 3. Find the number of terms in the series where a _{1} = 3, a_{2} = 6, a_{3 }= 12 a_{n} = 384**

**Options:**

**A. 8**

**B.9**

**C. 10**

**D. 7**

**Solution: ** In the given series,

a_{1} = 3,

r= 2,

a_{n} = 384

a_{n} = ar^{n-1}

384 = 3 * 2^{n-1}

128 = 2^{n-1}

2^{7 }= 2^{n-1}

n = 8

**Correct option: A**

**Type 3: Quickly Solve Sum of first ‘n’ terms of the series of GP: **S_{n}= [a_{1}(r^{n}-1)]/(r-1) and if r < 1 then S_{n}=[a_{1}(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 = 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

S_{n} = [a_{1}(r^{n}-1)] / (r-1)

S_{7} = [4(3^{7}-1)] / (3-1)

S_{7} = 4(2187-1) / 3-1

S_{7} = 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_{1}[(r_{n})-1]}/(r-1)

S_{6} = {2[(4^{6})-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**

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