Let the 1st term be a and common ratio be r, then
$a+ar+ar^2+.....=14$
$\implies\frac{a}{1-r}=14.........(1)$
And,
$a^3+(ar)^3+(ar^2)^3+.....=392$
$\implies\frac{a^3}{1-r^3}=392$
$\implies \frac{a^3}{(1-r)(1+r+r^2)}=392$
$\implies\frac{a}{1-r}*\frac{a^2}{1+r+r^2}=392$
$\implies\frac{a^2}{1+r+r^2}=392/14 [By (1)]$
$\implies\frac{a^2}{1+r+r^2}=28$
$\implies\frac{(14(1-r))^2}{1+r+r^2}=28$
By solving we get,
$2r^2-5r+2=0$
$\therefore r=2,\frac{1}{2}$
By (1),
$a=-14,7$
But, if we put a=-14, we would get $1+r+r^2+.....$ to be -1 which is never possible as r is positive.
$\therefore$ Correct answer is option (c) a=7
