The solution of the equation $(x + 1) + (x + 4) + (x + 7) + ......... + (x + 28) = 155$ is

Let $S_n$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20}=790$ and $S_{10}=145$, then $S_{15}-$ $S_5$ is:

The sums of $n$ terms of two arithmatic series are in the ratio $2n + 3:6n + 5$, then the ratio of their ${13^{th}}$ terms is

The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots,$ $b_{ m }$ is $2$ more than the common difference of $A.P.$ $a _{1}, a _{2}, \ldots, a _{ n } .$ If $a _{40}=-159, a _{100}=-399$ and $b _{100}= a _{70},$ then $b _{1}$ is equal to

If $\frac{1}{{b - c}},\;\frac{1}{{c - a}},\;\frac{1}{{a - b}}$ be consecutive terms of an $A.P.$, then ${(b - c)^2},\;{(c - a)^2},\;{(a - b)^2}$ will be in

For three positive integers $p , q , r , x ^{ pq p ^2}= y ^{ qr }= z ^{ p ^2 r }$ and $r=p q+1$ such that $3,3 \log _y x, 3 \log _z y, 7 \log _x z$ are in A.P. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to