The geometric series $a + ar + ar^2 + ar^3 +..... \infty$ has sum $7$ and the terms involving odd powers of $r$ has sum $'3'$, then the value of $(a^2 -r^2)$ is -

If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a $G.P.$ are $a, b$ and $c,$ respectively. Prove that

$a^{q-r} b^{r-p} c^{p-q}=1$

If the sum of the $n$ terms of $G.P.$ is $S$ product is $P$ and sum of their inverse is $R$, than ${P^2}$ is equal to

Given $a_1,a_2,a_3.....$ form an increasing geometric progression with common ratio $r$ such that $log_8a_1 + log_8a_2 +.....+ log_8a_{12} = 2014,$ then the number of ordered pairs of integers $(a_1, r)$ is equal to

The sum of first three terms of a $G.P.$ is $\frac{39}{10}$ and their product is $1 .$ Find the common ratio and the terms.

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0 .$ If $\alpha$ $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is