If the roots of the equation $x^3+3px^2+3qx-8=0$ are in an arithmetic progression,then $2p^3-3pq=$

Let $a_1, a_2, a_3, \ldots, a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i, 1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$,let $m=5n$. If $\frac{S_m}{S_n}$ does not depend on $n$,then $a_2$ is

$150$ 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

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ with $a_6 = 2$. Then the common difference of this $A.P.$,which maximizes the product $a_1 a_4 a_5$,is

The $p^{\text{th}}$,$q^{\text{th}}$,and $r^{\text{th}}$ terms of an $A.P.$ are $a$,$b$,and $c$ respectively. Show that $(q-r)a + (r-p)b + (p-q)c = 0$.

The number of terms in the series $101 + 99 + 97 + \dots + 47$ is