In an arithmetic progression,if $S_{40} = 1030$ and $S_{12} = 57$,then $S_{30} - S_{10}$ is equal to:

There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is

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

If the $n^{th}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$,then the ratio of their $12^{th}$ terms will be:

If $a_1, a_2, a_3, \dots, a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$,then the value of $\sum_{r=1}^{21} a_r$ is:

Show that the sum of $(m+n)^{th}$ and $(m-n)^{th}$ terms of an $A.P.$ is equal to twice the $m^{th}$ term.