Question

Find $10th$ and $16$th terms of the G.P. $256,128,64,........$

Answer 1

It is given that the first term of G.P is $a_1=256$ and the second term is $a_2=128$, therefore,

$r=\frac{a_2}{a_1}=\frac{128}{256}=\frac{1}{2}$

We know that the formula for the $n$th term of an G.P is $T_n=a{r}^{n-1}$, where $a$ is the first term, $r$ is the common ratio.

It is given that the first term is $a=256$ and the common ratio is $r=\frac{1}{2}$, thus,

$T_{10}=256\times {\left(\frac{1}{2}\right)}^{10-1}=256\times {\left(\frac{1}{2}\right)}^9=256\times \frac{1}{2^9}=256\times \frac{1}{512}=\frac{1}{2}$

$T_{16}=256\times {\left(\frac{1}{2}\right)}^{16-1}=256\times {\left(\frac{1}{2}\right)}^{15}=256\times \frac{1}{2^{15}}=256\times \frac{1}{32768}=\frac{1}{128}$

Hence, $T_{10}=\frac{1}{2}$ and $T_{16}=\frac{1}{128}$.