Question

The sum of the first three terms of an increasing geometric progression is $13$ and their product is $27$. Calculate the sum of the first five terms of the progression.

Answer 1

Let the first $3$ terms of $GP$ be $\frac{a}{r},a,ar$

$\therefore$ Product of $3$ terms$=27$

$\Rightarrow \frac{a}{r}\times a\times ar=27\Rightarrow a^3=27\Rightarrow a=3$

Also,$\frac{a}{r}+a+ar=13$

$\Rightarrow \frac{3}{r}+3+3r=13\Rightarrow \frac{3}{r}+3r+10\Rightarrow \frac{1}{r}+r=\frac{10}{3}\Rightarrow \frac{1+r^2}{r}=\frac{10}{3}\Rightarrow 3+3r^2=10r\Rightarrow 3r^2-10r+3=0\Rightarrow r=\frac{-(-10)\pm \sqrt{{(-10)}^2-4\times 3\times 3}}{2\times 3}=\frac{10\pm \sqrt{100-36}}{6}=\frac{10\pm \sqrt{64}}{6}r=\frac{10\pm 8}{6}=\frac{18}{6},\frac{2}{6}=3,\frac{1}{3}$

Taking $r=3$, as it is an incrasing $GP$.

$S_5=\frac{a(1-r^5)}{1-r}=\frac{3(1-3^5)}{1-3}=\frac{3\times (1-243)}{(-2)}=\frac{3\times (-242)}{(-2)}{S}_5=363$