Question

The 4th term of a geometric progression is 2/3 and the seventh term is 16/81. Find the geometric series.

Answer 1

In a geometric series, if the fourth term is 2/3 and seventh term is 16:81 , then what is the first term of the series?

A) 2/ 3

B) 4/ 9

C) 8/ 27

D) 9/ 4

Correct Answer:

D) 9/ 4

Answer 2

Given,

In a Geometric Progression,

$a4 = \frac{2}{3} \\ a. {r}^{3} = \frac{2}{3} (equation \: 1)$

$a7 = \frac{16}{81} \\ a. {r}^{6} = \frac{16}{81} (equation \: 2)$

$\frac{equation \: 1}{equation \: 2} = \frac{a. {r}^{6} }{a. {r}^{3} } = \frac{ \frac{16}{81} }{ \frac{2}{3} }$

${r}^{6 - 3} = \frac{16}{81} \times \frac{3}{2} = \frac{8}{27}$

${r}^{3} = ({ \frac{2}{3} })^{3}$

$r = \frac{2}{3}$

Now substituting r = 2/3 in equation 1 we get,

$a( { \frac{2}{3} })^{3} = \frac{2}{3}$

$a = \frac{2}{3} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}$

$a = \frac{9}{4}$

Therefore the G.P is a,ar,ar^2,ar^3,.....

$\frac{9}{4} , \frac{9}{4} \times \frac{2}{3} , \frac{9}{4} \times ({ \frac{2}{3} })^{2} .....$

Therefore the Geometric progression is

$\frac{9}{4} , \frac{3}{2} ,1, ...$

Additional information:

what is a Geometric progression?

The successive terms are obtained by multiplying the preceding term by a fixed number. Such type of list of numbers is said to form Geometric Progression.

1. The fixed number is called the common ration 'r' of GP.

2. The general form of GP is a,ar,ar^2,ar^3...

Example: 30,90,270,810,....