Let $\mathrm{a}$ and $\mathrm{b}$ be be two distinct positive real numbers. Let $11^{\text {th }}$ term of a $GP$, whose first term is $a$ and third term is $b$, is equal to $p^{\text {th }}$ term of another $GP$, whose first term is $a$ and fifth term is $b$. Then $\mathrm{p}$ is equal to

Let $M=2^{30}-2^{15}+1$, and $M^2$ be expressed in base $2$.The number of $1$'s in this base $2$ representation of $M^2$ is

If $a,\,b,\,c$ are in $G.P.$, then

Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+k, \in N$. If $c_2=5$ and $c_3=13 / 4$ then $\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)$ is equal to

If $a, b, c$ and $d$ are in $G.P.$ show that:

$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$

Let $b_1, b_2,......, b_n$ be a geometric sequence such that $b_1 + b_2 = 1$ and $\sum\limits_{k = 1}^\infty  {{b_k} = 2} $ Given that $b_2 < 0$ , then the value of $b_1$ is