Mathematics - Sequence & Series Question with Solution | TestHub

Let the A.P. be$a, a+d, a+2d, .....$

Given,$\left(a+d\right).\left(a+8d\right)={\left(a+4d\right)}^2$

$a^2+9ad+8d^2=a^2+8ad+16d^2$

$8d^2-ad=0$

$d\left[8d-a\right]=0$

$\because$$d \ne 0$

$d=\frac{a}{8}$

So,$2^{nd}$term,$5^{th}$term,$9^{th}$term

would be$\left(a+\frac{a}{8}\right)$,$\left(a+\frac{a}{2}\right)$,$\left(a+a\right)$

$\frac{9a}{8}$,$\frac{3a}{2}$,$2a$

Common Ratio$=\frac{2^{nd} term}{1^{st} term}=\frac{3a\times 8}{2\times 9a} =\frac{4}{3}$