Question

the fourth term of a GP is 2/3 and 7th term is 16/81 find geometric series

Answer 1

hope it helps.... .. ,
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Answer 2

Answer:

The required GP is

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

Step-by-step explanation:

It is given that fourth term of a GP is 2/3 and 7th term is 16/81.

Let the first term of the GP be a and the common ratio is r.

The nth term of a GP is

$a_n=ar^{n-1}$

4th term of the GP is 2/3.

$a_4=ar^{4-1}$

$\frac{2}{3}=ar^{3}$                      .... (1)

7th term of the GP is 16/81..

$a_7=ar^{7-1}$

$\frac{16}{81}=ar^{6}$                  .... (2)

Divide equation (2) by equation (1).

$\frac{3}{2}\times \frac{16}{81}=r^{3}$

$\frac{8}{27}=r^{3}$

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

The common ratio is 2/3.

Put this value in equation (1).

$\frac{2}{3}=a(\frac{2}{3})^3$  

$\frac{2}{3}=a(\frac{8}{27})$  

$a=\frac{2}{3}\times \frac{27}{8}=\frac{9}{4}$

The first term is 9/4.

The required GP is

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

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