The  sum of first 24 terms of the A.P, $a_1,a_2,a_3,\mathrm{....}$ if it is known that  $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$

We known that in an A.P the sum of the terms

equidistant  from the beginning and end is always same and is equal to

the sum of first and last term, ie. $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\mathrm{.....}$

So if an A.P. consists  of 24 terms, then 

$\begin{array}{l}{a}_1+a_{24}=a_3+a_{20}=a_{10}+a_{15}\\ a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225\\ \Rightarrow a_1+a_{24}=\frac{225}{3}=75\end{array}$

$\begin{array}{l}\therefore S_{24}=\frac{24}{2}\left(a_1+a_{24}\right)\left[u\sin g{S}_n=\frac{n}{2}\left(a_1+a_n\right)\right]\\ =12\left(75\right)\\ =900\left(\text{ using }\right(i\left)\right)\end{array}$