The first term of an $A.P. $ is $2$ and common difference is $4$. The sum of its $40$ terms will be

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $\mathrm{S}_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $S_{15}-S_5$ is equal to:

Write the first five terms of the following sequence and obtain the corresponding series :

$a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n\,>\,2$

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 $a$, $b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x ^{2}-8 ax +2 a =0$ and $q$ and $s$ are the roots of the equation $x^{2}+12 b x+6 b$ $=0$, such that $\frac{1}{ p }, \frac{1}{ q }, \frac{1}{ r }, \frac{1}{ s }$ are in A.P., then $a ^{-1}- b ^{-1}$ is equal to $......$

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$.. If $S_4 = 16$ and $S_6 = -48$, then $S_{10}$ is equal to