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

Answer: 1

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