Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{px}^2+\mathrm{qx}-$ $r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non-constant G.P and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^2$ is :

If $a, b, c, d$ and $p$ are different real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right)\, \leq \,0,$ then show that $a, b, c$ and $d$ are in $G.P.$

If in a $G.P.$ of $64$ terms, the sum of all the terms is $7$ times the sum of the odd terms of the $G.P,$ then the common ratio of the $G.P$. is equal to

The sum of few terms of any ratio series is $728$, if common ratio is $3$ and last term is $486$, then first term of series will be

If $a,\;b,\;c,\;d$ and $p$ are different real numbers such that $({a^2} + {b^2} + {c^2}){p^2} - 2(ab + bc + cd)p + ({b^2} + {c^2} + {d^2}) \le 0$, then $a,\;b,\;c,\;d$ are in

The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a $G.P.$ are $p, q$ and $s,$ respectively. Show that $q^{2}=p s$