If a+b+...+l is a GP, prove that its sum is bl-a^2/b-a

given: a,b, ........I are in GP.
therefore 
common ratio r = b/a
let the number of terms in GP be n.
therefore the sum of n terms of GP is $\frac{a(r^n-1)}{r-1}$ [since a is the first term of GP]
nth term of GP is $I=a.r^{n-1}\Rightarrow r^{n-1}=\frac{I}{a}$
$$\begin{gathered} therefore r^n=r.r^{n-1}=r.\frac{I}{a} \\ {r}^n=\frac{b}{a}.\frac{I}{a}=\frac{bI}{a^2} \end{gathered}$$
sum $=\frac{a\left({\displaystyle \frac{bI}{a^2}}-1\right)}{{\displaystyle \frac{b}{a}}-1}=\frac{a.\left({\displaystyle \frac{bI-a^2}{a^2}}\right)}{{\displaystyle \frac{b-a}{a}}}$
$=\frac{bI-a^2}{b-a}$

hope this helps you