If $a_1, a_2, a_3, \dots, a_n$ are in $H.P.$, then the expression $a_1 a_2 + a_2 a_3 + \dots + a_{n-1} a_n$ is equal to:

The product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ is

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$,where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

If $a$ is the first term of a $G.P.$,$l$ is the $n^{th}$ term,and $P$ is the product of the first $n$ terms,then $P=$

The sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is

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