Question

If $T_1,T_2,T_7$ an $A.P$ constitute a $G.P$. Whose sum is $93$, then find the numbers.

Answer 1

$a,a+d,a+6d$ are in $G.P.$
$\therefore {(a+d)}^2=a(a+6d)\therefore d^2-4ad=0$
$d(d-4a)=0\therefore d=0$ or $d=4a$
Since $d$ cannot be zero, therefore we take $d=4a$.
$\therefore$ Numbers are $a,5a,25a$ where$a+5a+25a=31a=93$ given $\therefore a=3,d=12$.
Hence the numbers are $3,15,75$ which are clearly in $G.P.$