Mathematics - Sequence & Series Question with Solution | TestHub

Let 3 numbers in AP are $a,(a+6),(a+12)$ Also, first and last number out of 4 numbers are equal $\therefore 4$ numbers are $(a+12),a,(a+6),(a+12)$ Also, given, first 3 numbers are in GP $\begin{array}{ll}\Rightarrow & a^2=(a+12)(a+6)\\ \Rightarrow & a^2=a^2+18a+72\\ \Rightarrow & a=-\frac{72}{18}=-4\end{array}$ $\therefore 4$ numbers are $8,-4,2,8$