Question

Let $a_n$ denote the $n^{\text{th}}$ term of a geometric progression with common ratio less than $1.$ If $a_1+a_2+a_3=13$ and $a_1^2+a_2^2+a_3^2=91,$ then the value of $a_{10}$ is

• A. $3^{10}$
• B. $3^{11}$
• C. $\frac{1}{3^{10}}$
• D. $\frac{1}{3^7}$

Answer 1

The correct option is D $\frac{1}{3^7}$

Let $a_1=\frac{a}{r},a_2=a,a_3=ar$
$a_1+a_2+a_3=13$
$\Rightarrow \frac{a}{r}+a+ar=13$
$\Rightarrow a(\frac{1}{r}+r)=13-a$
$\Rightarrow \frac{1}{r}+r=\frac{13-a}{a}\cdots \left(1\right)$

$a_1^2+a_2^2+a_3^2=91$
$\Rightarrow a^2(\frac{1}{r^2}+1+r^2)=91$
$\Rightarrow a^2[{(\frac{1}{r}+r)}^2-1]=91$
$\Rightarrow a^2[{\left(\frac{13-a}{a}\right)}^2-1]=91\left[\text{From}\right(1\left)\right]$
$\Rightarrow (13-a{)}^2-a^2=91$
$\Rightarrow a=3$

So, $\frac{1}{r}+r=\frac{10}{3}$
$\Rightarrow 3r^2-10r+3=0$
$\Rightarrow r=\frac{1}{3}[\because r<1]$

$a_1=9\therefore a_{10}=9{\left(\frac{1}{3}\right)}^9={\left(\frac{1}{3}\right)}^7$