Mathematics - Sequence & Series Question with Solution | TestHub

Let $a,b,c,d$ be four numbers of the sequence. Now, according to the question $b^2=ac$ and $c-b=6$ and $a-c=6$ Also, given $a=d$ $$\begin{array}{r}\therefore b^2=ac\Rightarrow b^2=a\left[\frac{a+b}{2}\right]\\ (\because 2c=a+b)\end{array}$$ $\Rightarrow a^2-2b^2+ab=0$ Now, $c-b=6$ and $a-c=6$, gives $a-b=12$ $\Rightarrow b=a-12$ $\therefore a^2-2b^2+ab=0$ $\Rightarrow a^2-2(a-12{)}^2+a(a-12)=0$ $\Rightarrow a^2-2a^2-288+48a+a^2-12a=0$ $\Rightarrow 36a=288\Rightarrow a=8$