Question

The first term of a geometric series is $375$ and the fourth term is $192$. Find the common ratio and the sum of the first $14$ terms

Answer 1

Let $a$ be the first term and $r$ be the common ratio of the given G.P.
Given that $a=375,t_4=192$.
Now, $t_n=a{r}^{n-1}$
Therefore, $t_4=375r^3\Rightarrow 375r^3=192$
$\Rightarrow r^3=\frac{192}{375}\Rightarrow r^3=\frac{64}{125}$
$\Rightarrow r^3={\left(\frac{4}{5}\right)}^3\Rightarrow r=\frac{4}{5}$, which is the required common ratio.
Now, $S_n=a\left[\frac{r^n-1}{r-1}\right]$ if $r\ne 1$
Thus, $S_{14}=\frac{275[{\left(\frac{4}{5}\right)}^{14}-1]}{\frac{4}{5}-1}=(-1)\times 5\times 375[{\left(\frac{4}{5}\right)}^{14}-1]$
$=\left(375\right)\left(5\right)[1-{\left(\frac{4}{5}\right)}^{14}]=1875[1-{\left(\frac{4}{5}\right)}^{14}]$.