Three positive numbers form an increasing $G.P.$ If the middle term in this $G.P.$ is doubled,the new numbers are in $A.P.$ then the common ratio of the $G.P.$ is:

Let $S_k = \frac{1 + 2 + 3 + .... + k}{k}$. If $S_1^2 + S_2^2 + ....... + S_{10}^2 = \frac{5}{12}A$,then $A$ is equal to

For $x \in R$,let $[x]$ denote the greatest integer $\le x$. Find the sum of the series $\left[ -\frac{1}{3} \right] + \left[ -\frac{1}{3} - \frac{1}{100} \right] + \left[ -\frac{1}{3} - \frac{2}{100} \right] + \dots + \left[ -\frac{1}{3} - \frac{99}{100} \right]$.

If ${S_n} = nP + \frac{1}{2}n(n - 1)Q$,where ${S_n}$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

If $a, b, c$ are the $p^{th}, q^{th}$ and $r^{th}$ terms of a $G.P.$,then $\left( \frac{c}{b} \right)^p \left( \frac{b}{a} \right)^r \left( \frac{a}{c} \right)^q$ is equal to

If $a_1, a_2, a_3, \dots$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is