Question

Find the G.P. in which the 10th term is 320 and the 6th term is 20.

Answer 1

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 $10$th term is $T_{10}=320$, therefore,

$T_{10}=a{r}^{10-1}\Rightarrow 320=a{r}^9\Rightarrow a=\frac{320}{r^9}.....\left(1\right)$

It is also given that the $6$th term is $T_6=20$, therefore using equation 1,

$T_6=a{r}^{6-1}\Rightarrow 20=\frac{320}{r^9}\times r^5\Rightarrow \frac{20}{320}=\frac{1}{r^4}\Rightarrow \frac{1}{r^4}=\frac{1}{16}\Rightarrow \frac{1}{r^4}=\frac{1}{2^4}\Rightarrow r^4=2^4\Rightarrow r=2$

Now, substitute the value of $r$ in equation 1:

$a=\frac{320}{2^9}=\frac{320}{512}=\frac{5}{8}$

Therefore, the terms of G.P are:

$a_1=\frac{5}{8}$

$a_2=a_1\times r=\frac{5}{8}\times 2=\frac{5}{4}$

$a_3=a_2\times r=\frac{5}{4}\times 2=\frac{5}{2}$

$a_4=a_3\times r=\frac{5}{2}\times 2=5$

Hence, the G.P is $\frac{5}{8},\frac{5}{4},\frac{5}{2},5,.....$.