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 :-

In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$

If $p$ times the ${p^{th}}$ term of an $A.P.$ is equal to $q$ times the ${q^{th}}$ term of an $A.P.$, then ${(p + q)^{th}}$ term is

If $a,\;b,\;c$ are in $A.P.$, then $\frac{1}{{bc}},\;\frac{1}{{ca}},\;\frac{1}{{ab}}$ will be in

${7^{th}}$ term of an $A.P.$ is $40$, then the sum of first $13$ terms is

If the $9^{th}$ term of an $A.P.$ be zero, then the ratio of its $29^{th}$ and $19^{th}$ term is