The first term of an $A.P.$ of consecutive integers is ${p^2} + 1$ The sum of $(2p + 1)$ terms of this series can be expressed as
${(p + 1)^2}$
${(p + 1)^3}$
$(2p + 1){(p + 1)^2}$
${p^3} + {(p + 1)^3}$
Let ${S_n}$ denotes the sum of $n$ terms of an $A.P.$ If ${S_{2n}} = 3{S_n}$, then ratio $\frac{{{S_{3n}}}}{{{S_n}}} = $
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
If the sum of the first $2n$ terms of $2,\,5,\,8...$ is equal to the sum of the first $n$ terms of $57,\,59,\,61...$, then $n$ is equal to
If $\log 2,\;\log ({2^n} - 1)$ and $\log ({2^n} + 3)$ are in $A.P.$, then $n =$
The sum of $n$ terms of two arithmetic progressions are in the ratio $(3 n+8):(7 n+15) .$ Find the ratio of their $12^{\text {th }}$ terms.