- Prepare
All Platforms Programming Aptitude Syllabus Interview Preparation Interview Exp. Off Campus - Prime Video
- Prime Mock

- Interview Experience
- Prime VideoNew
- Prime Mock
- Interview Prep
- Nano Degree
- Prime Video
- Prime Mock

# 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, ar ^{2}, ar^{3}**, …….

where

**a** = the first term

**r** = the common ratio.

**Type 1: How To Solve Geometric Progression Quickly for ****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} = a_{1}r^{n-1}

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

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

a_{3} = 45 x 0.04

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

** D. n**

**Solution: **In the given series,

a = \frac{- 1}{2}

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

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

2r^{2} – 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: **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 = \frac{3}{1} = 3,

a_{n }=729

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

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

729 = 3^{n-1}

3^{6} = 3^{n-1}

6 = n-1

n = 7

**Correct option: D**

**Question 2. 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 x 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: 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}

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 \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**

**Get over 200+ course One Subscription**

Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others

- AP GP HP – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Arithmetic Progressions – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Harmonic Progressions – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts

Login/Signup to comment