Question

Find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.

Answer 1

Let the three numbers of G.P. be $a,ar,a{r}^2$.

Given that the sum of three numbers is $\frac{13}{3}$, i.e.,

$a+ar+a{r}^2=\frac{13}{3}$

$\Rightarrow a(1+r+r^2)=\frac{13}{3}.....\left(1\right)$

Also given that the sum of their squares is $\frac{91}{9}$

$a^2+a^2{r}^2+a^2{r}^4=\frac{91}{9}$

$\Rightarrow a^2(1+r^2+r^4)=\frac{91}{9}.....\left(2\right)$

Now, from ${eq}^n\left(1\right)$, i.e.,

$a(1+r+r^2)=\frac{13}{3}$

Squaring both sides, we get

$a^2(1+r^2+r^4+(2r+2r^3+2r^2))=\frac{169}{9}.....\left(3\right)$

Subtracting ${eq}^n\left(2\right)$ from $\left(3\right)$, we have

$a^2(2r+2r^3+2r^2)=\frac{169}{9}-\frac{91}{9}$

$2ar.\left(a(1+r+r^2)\right)=\frac{78}{9}$

$2ar\left(\frac{13}{3}\right)=\frac{78}{9}\left[\text{From}\left(1\right)\right]$

$\Rightarrow ar=1$

$\Rightarrow r=\frac{1}{a}.....\left(4\right)$

Substituting the value of $r$ in ${eq}^n\left(1\right)$, we have

$a+1+\frac{1}{a}=\frac{13}{3}$

$\Rightarrow 3a^2+3a+3=13a$

$\Rightarrow 3a^2-10a+3=0$

$\Rightarrow (a-3)(3a-1)=0$

$\Rightarrow a=3,\frac{1}{3}$

Case I:-

If $a=\frac{1}{3}$

$\Rightarrow r=3$

$\therefore$ The numbers will be= $\frac{1}{3},1,9$.

Case II:-

If $a=3$

$\Rightarrow r=\frac{1}{a}$

$\therefore$ The numbers will be= $3,1,\frac{1}{3}$