Mathematics - Sequence & Series Question with Solution | TestHub

Let the first terms of A.P. be$A$and common difference be$d,$then

$a=A+6d$

$b=A+10d$

$c=A+12d$

Given$a, b, c$are in G. P.$\Rightarrow b^2=ac$

$\Rightarrow {\left(A+10d\right)}^2=\left(A+6d\right)\left(A+12d\right)$

$\Rightarrow A^2+100d^2+20Ad=A^2+18Ad+72d^2$

$\Rightarrow A=-14d$

Now$\frac{a}{c}=\frac{A+6d}{A+12d}=\frac{-14d+6d}{-14d+12d}=\frac{-8d}{-2d}=4$