Check everything!

As they are in G.P,

$\left( \dfrac{a+n}{b+n} \right)^2 = \dfrac{a}{b} * \left (\dfrac{a+2n}{b+2n} \right)$

Expand it,

$\dfrac{a^2+n^2+2an}{b^+n^2+2bn} = \dfrac{a^2+2an}{b^2+2bn}$

$1+ \dfrac{n^2}{a^2+2an} = 1+\dfrac{n^2}{b^2+2bn}$

$a^2+2an - b^2-2bn = 0$

$(a-b)(a+b+2n) =0$ ,

Now if $a≠b \implies a+b=-2n$ and if $a=b$ , then $a+b$ has infinitely many solutions.

Now , they are also in A.P, so,

$2\left(\dfrac{a+n}{b+n} \right) = \dfrac{a}{b} + \left (\dfrac{a+2n}{b+2n} \right)$

$ab^2+2abn+nb^2+2n^2b=ab^2+2abn+an^2+b^2n+bn^2 \implies n^2(a-b) = 0$

As $n≠0$ $\implies$ $a=b$ . Also if they are in H.P we will get the same result due to symmetry in ratios.

But if $a=b$ , then there are infinite values of $a+b$