$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day, $4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

If ${a_1},\,{a_2},....,{a_{n + 1}}$ are in $A.P.$, then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ..... + \frac{1}{{{a_n}{a_{n + 1}}}}$ is

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

Let $x _1, x _2 \ldots ., x _{100}$ be in an arithmetic progression, with $x _1=2$ and their mean equal to $200$ . If $y_i=i\left(x_i-i\right), 1 \leq i \leq 100$, then the mean of $y _1, y _2$, $y _{100}$ is

Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals

If the first term of an $A.P.$ is $3$ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms, then the common difference of this $A.P.$ is :