The sixth term of an $A.P.$ is equal to $2$, the value of the common difference of the $A.P.$ which makes the product ${a_1}{a_4}{a_5}$ least is given by

Let the sum of $n, 2 n, 3 n$ terms of an $A.P.$ be $S_{1}, S_{2}$ and $S_{3},$ respectively, show that $S_{3}=3\left(S_{2}-S_{1}\right)$

The $20^{\text {th }}$ term from the end of the progression $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots .,-129 \frac{1}{4}$ is :-

If in the equation $a{x^2} + bx + c = 0,$ the sum of roots is equal to sum of square of their reciprocals, then $\frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in

In an $A.P.,$ if $p^{\text {th }}$ term is $\frac{1}{q}$ and $q^{\text {th }}$ term is $\frac{1}{p},$ prove that the sum of first $p q$ terms is $\frac{1}{2}(p q+1),$ where $p \neq q$

If the first term of an $A.P. $ be $10$, last term is $50$ and the sum of all the terms is $300$, then the number of terms are