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
$\frac{{41}}{{11}}$
$\frac{7}{2}$
$\frac{2}{7}$
$\frac{{11}}{{41}}$
Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals
If $\tan \,n\theta = \tan m\theta $, then the different values of $\theta $ will be in
The sixth term of an $A.P.$ is equal to $2$, the value of the common difference of the $A.P.$ which makes the product ${a_1}{a_4}{a_5}$ least is given by
If the sum of first $n$ terms of an $A.P.$ be equal to the sum of its first $m$ terms, $(m \ne n)$, then the sum of its first $(m + n)$ terms will be
If $a,\,b,\,c$ are in $A.P.$, then $(a + 2b - c)$ $(2b + c - a)$ $(c + a - b)$ equals