Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$ If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals
$\frac{{41}}{{11}}$
$\frac{7}{2}$
$\frac{2}{7}$
$\frac{{11}}{{41}}$
If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.
The sum of the integers from $1$ to $100$ which are not divisible by $3$ or $5$ is
If $x_1 , x_2 , ..... , x_n$ and $\frac{1}{{{h_1}}},\frac{1}{{{h^2}}},......\frac{1}{{{h_n}}}$ are two $A.P' s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$, then $x_5. h_{10}$ equals
If the sum of the first $n$ terms of the series $\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 $ , then $n$ equals
The difference between an integer and its cube is divisible by