Question

Find the 5th term of the G. P.: 1/7,1/14, 1/28 ___

Answer 1

The 5th term of the Geometric Progression is $\frac{1}{112}$

Step-by-step explanation:

Given G.P. series is : $\frac{1}{7} , \frac{1}{14} , \frac{1}{28}$ __

The nth term of the Geometric Progression A(n) is given by

=> $A\left ( n \right ) = a\times r^{n-1}$

Where, a is the 1st term of an G.P.

            r is the common ratio of an G.P.

            n is the number of terms

So, here $a = \frac{1}{7}$

=> $r = \frac{a2}{a1}$

    $r = \frac{\frac{1}{14}}{\frac{1}{7}}$

    $r = \frac{1}{2}$

=> For 5th term of the G.P. , put n = 5 in the nth term equation of an G.P.

=> $A\left ( n=5 \right ) = a\times r^{5-1}$

    $A\left ( n=5 \right ) = \frac{1}{7}\times \left ( \frac{1}{2} \right )^{5-1}$

    $A\left ( n=5 \right ) = \frac{1}{7}\times \left ( \frac{1}{2} \right )^{4}$

    $A\left ( n=5 \right ) = \frac{1}{7}\times \left ( \frac{1}{16} \right )$

    $A\left ( n=5 \right ) = \frac{1}{112}$

=> The 5th term of the Geometric Progression is $\frac{1}{112}$

Answer 2

1/112

Given:

A G.P:  1/7,1/14, 1/28 ___

First term of the GP(a) = 1/7

Common Ratio of the GP(r) = 1/2

Formula used to solve these type of questions:

= Tn = ar^n-1

Substituting all the values that are known to us in this equation we get:

= (1/7)(1/2)^5 - 1

= 1/7 x 1/2^4

= 1/7 x 1/16

= 1/112

Therefore, the 5 th term of the Geometric Progression is 1/112.