Question

If the sum of an infinite geometric series is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$, then find the first term of the series.

Answer 1

Let the first term be $a$ and common ratio be $r$.
$a+a r+a r^{2}+\ldots=3$
$\Rightarrow \frac{a}{1-r}=3 \,\,\,...(i)$
$a^{3}+a^{3} r^{3}+a^{3} r^{6}+\ldots=\frac{27}{19} $
$\Rightarrow \frac{a^{3}}{1-r^{3}}=\frac{27}{19}\,\,\,...(ii)$
From (i) and (ii), we get
$\left(\frac{a}{1-r}\right)^{3} \div\left(\frac{a^{3}}{1-r^{3}}\right)$
$=\left(3^{3}\right) \div \frac{27}{19}$
$\Rightarrow \frac{1-r^{3}}{(1-r)^{3}}=19$
$\Rightarrow \frac{1+r+r^{2}}{(1-r)^{2}}=19$
$\Rightarrow 1+r+r^{2}=19-38 r+19 r^{2}$
$\Rightarrow 18 r^{2}-39 r+18=0$
$\Rightarrow 6 r^{2}-13 r+6=0$
$\Rightarrow(2 r-3)(3 r-2)=0$
$\Rightarrow r=\frac{2}{3}\,\,\ldots[$ As $|r|< 1]$
$\frac{a}{1-\frac{2}{3}}=3\,\,\,\ldots$ [From (i)]
$\Rightarrow a=1$