How To Solve Geometric Progression Questions Quickly

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