${7^{th}}$ term of an $A.P.$ is $40$, then the sum of first $13$ terms is

If the sum of three numbers of a arithmetic sequence is $15$ and the sum of their squares is $83$, then the numbers are

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

If $a_1, a_2, a_3, …….$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is

Let $AP ( a ; d )$ denote the set of all the terms of an infinite arithmetic progression with first term a and common difference $d >0$. If $\operatorname{AP}(1 ; 3) \cap \operatorname{AP}(2 ; 5) \cap \operatorname{AP}(3 ; 7)=\operatorname{AP}( a ; d )$ then $a + d$ equals. . . . .

If $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$ be the $A.M.$ of $a$ and $b$, then $n=$