Question

Find out three numbers which are in G. P. such that their summation is $13$ and the sum of their squares is $91$

• A. $3,1,9$
• B. $1,3,9$
• C. $1,9,3$
• D. $\frac{13}{3},\frac{13}{3},\frac{13}{3}$

Answer 1

The correct option is B

$1,3,9$

Step 1. Let the three numbers be $a,ar,a{r}^2$

According to the question,

$a+ar+a{r}^2=13$

$\Rightarrow$ $a(1+r+r^2)=13$ ……(1)

And their sum $a^2+a^2{r}^2+a^2{r}^4=91$

$\Rightarrow$$a^2\left(1+r^2+r^4\right)=91$ ……(2)

Step 2. squaring equation (1) and divide it by equation (2)

$\frac{a^2{\left(1+r+r^2\right)}^2}{a^2(1+r^2+r^4)}=\frac{169}{91}$

$\Rightarrow$ $\frac{{\left(1+r+r^2\right)}^2}{{(1+r^2)}^2-r^2}=\frac{13}{7}$

$\Rightarrow$$\frac{{\left(1+r+r^2\right)}^2}{(1+r^2+r\left)\right(1+r^2-r)}=\frac{13}{7}$

$\Rightarrow$ $\frac{\left(1+r+r^2\right)}{\left(1+r^2-r\right)}=\frac{13}{7}$

$\Rightarrow$ $7\left(1+r+r^2\right)=13\left(1+r^2-r\right)$

$\Rightarrow$ $\left(7+7r+7r^2\right)=\left(13+13r^2-13r\right)$

$\Rightarrow$ $6r^2-20r+6=0$

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

$\Rightarrow$ $3r^2-9r-r+3=0$

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

$\Rightarrow$ $\left(3r-1\right)\left(r-3\right)=0$

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

Step 3. Substitute value of $r$ in equation (1)

$a\left(1+3+9\right)=13$and $a\left(1+\frac{1}{3}+\frac{1}{9}\right)=13$

$13a=13$and $\frac{13a}{9}=13$

$a=1$ and $a=9$

Now, numbers are $a,ar,a{r}^2$

If $r=3$ and $a=1$

then, the numbers are $1,3,9$

If $r=\frac{1}{3}$ and $a=9$

then, the numbers are $9,3,1$

Hence, Option ‘B’ is Correct.