If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

Find the $10^{th}$ term of the geometric series $5+25+125+\ldots$

$\sqrt {\underbrace {111........1}_{200\,digits} - \underbrace {222.......2}_{100\,digits}} $ equals :-

Let $S_n = 1 + q + q^2 + ..... + q^n$ and $T_n = 1 + \left( \frac{q + 1}{2} \right) + \left( \frac{q + 1}{2} \right)^2 + ...... + \left( \frac{q + 1}{2} \right)^n$ where $q$ is a real number and $q \neq 1$. If $^{101}C_1 + ^{101}C_2 \cdot S_1 + ...... + ^{101}C_{101} \cdot S_{100} = \alpha \cdot T_{100}$,then $\alpha$ is equal to

The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive,then its $7^{th}$ term is

If the sum $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ up to $20$ terms is equal to $\frac{k}{21}$,then $k$ is equal to