The sums of $n$ terms of three $A.P.'s$ whose first term is $1$ and common differences are $1, 2, 3$ are ${S_1},\;{S_2},\;{S_3}$ respectively. The true relation is

The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is

Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$

Let $s _1, s _2, s _3, \ldots \ldots, s _{10}$ respectively be the sum to 12 terms of 10 A.P.s whose first terms are $1,2,3, \ldots, 10$ and the common differences are $1,3,5, \ldots, 19$ respectively. Then $\sum \limits_{i=1}^{10} s _{ i }$ 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. . . . . 

If the ${p^{th}}$ term of an $A.P.$ be $\frac{1}{q}$ and ${q^{th}}$ term be $\frac{1}{p}$, then the sum of its $p{q^{th}}$ terms will be