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. [4 Marks]

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, $t_n=a{r}^{n-1}$

$\therefore t_4=375r^3\Rightarrow 375r^3=192$ [1 Mark]

$\Rightarrow r^3=\frac{192}{375}\Rightarrow r^3=\frac{64}{125}$
$\Rightarrow r^3={\left(\frac{4}{5}\right)}^3$
$\Rightarrow r=\frac{4}{5}$ [1 Mark]

Now, $S_n=a\left[\frac{r^n-1}{r-1}\right]$ if $r\ne 1$

$S_{14}=\frac{375[{\left(\frac{4}{5}\right)}^{14}-1]}{\frac{4}{5}-1}$ [1 Mark]
$=(-1)\times 5\times 375[{\left(\frac{4}{5}\right)}^{14}-1]$

$=\left(375\right)\left(5\right)[1-{\left(\frac{4}{5}\right)}^{14}]$
$=1875[1-{\left(\frac{4}{5}\right)}^{14}]$ [1 Mark]