The formula for the sum of infinite geometric sequence is given by
$$1 + x + x^2 + x^3 + \cdots \cdots = \frac{1}{1-x}$$
where $\mid x \mid <1$
Now
$$l=\frac{1}{1-a}$$
$$m=\frac{1}{1-b}$$
$$n=\frac{1}{1-c}$$
since $a$,$b$,$c$ are in AP let initial term is $s$ and common difference is $d$ then $a=s$, $b=s+d$, $c=s+2d$
$\frac{1}{l} =1-a=1-s $
$\frac{1}{m} =1-b=1-s-d $
$\frac{1}{n} =1-c=1-s-2d $
clearly $\frac{1}{l},\frac{1}{m},\frac{1}{n}$ are in A.P, with the common difference $-d$
$\implies l,m,n $ are in H.P.
The answer is OPTION C