Find the $25^{th}$ common term of the following $A.P.'s$

$S_1 = 1, 6, 11, .....$

$S_2 = 3, 7, 11, .....$

If the sum of first $n$ terms of an $A.P.$ is $c n^2$, then the sum of squares of these $n$ terms is

Let $a_1 , a_2, a_3, .... , a_n$, be in $A.P$. If $a_3 + a_7 + a_{11} + a_{15} = 72$ , then the sum of its first $17$ terms is equal to

Let $l_1, l_2, \ldots, l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_1$, and let $w_1, w_2, \ldots, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_2$, where $d_1 d_2=10$. For each $i=1,2, \ldots, 100$, let $R_i$ be a rectangle with length $l_i$, width $w_i$ and area $A_i$. If $A_{51}-A_{50}=1000$, then the value of $A_{100}-A_{90}$ is. . . . . 

Given an $A.P.$ whose terms are all positive integers. The sum of its first nine terms is greater than $200$ and less than $220$. If the second term in it is $12$, then its $4^{th}$ term is

The $8^{\text {th }}$ common term of the series $S _1=3+7+11+15+19+\ldots . .$ ; $S _2=1+6+11+16+21+\ldots .$ is $.......$.