The sum of $1 + 3 + 5 + 7 + .........$ upto $n$ terms is
${(n + 1)^2}$
${(2n)^2}$
${n^2}$
${(n - 1)^2}$
If $a_m$ denotes the mth term of an $A.P.$ then $a_m$ =
Let $a_1, a_2 , a_3,.....$ be an $A.P$, such that $\frac{{{a_1} + {a_2} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + ..... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}};p \ne q$. Then $\frac{{{a_6}}}{{{a_{21}}}}$ is equal to
Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is square of an integer. The least possible value of the number of digits of $c$ is
If the sum of first $n$ terms of an $A.P.$ is $cn(n -1)$ , where $c \neq 0$ , then sum of the squares of these terms is
If the $A.M.$ between $p^{th}$ and $q^{th}$ terms of an $A.P.$ is equal to the $A.M.$ between $r^{th}$ and $s^{th}$ terms of the same $A.P.$, then $p + q$ is equal to