The sum of first $n$ natural numbers is

Sum of the first $p, q$ and $r$ terms of an $A.P.$ are $a, b$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

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 $20^{\text {th }}$ 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 :-

Find the $20^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=\frac{n(n-2)}{n+3}$ 

Find the sum to $n$ terms of the $A.P.,$ whose $k^{\text {th }}$ term is $5 k+1$