Find the sum of first 24 terms of the A.P. `a_1, a_2, a_3, ,` if it is know that `a_1+a_5+a_(10)+a_(15)+a_(20)+a_(24)=225.`

Explanation

We know that in an A.P. the sum of the terms equidistant from the begaining and end is always same and is equal to the sum of the first and last terms, i.e., `a_(1)+a_(n)=a_(2)+a_(n-1)=a_(3)+a_(n-1)=cdotcdotcdot`.
So, if an A.P consists of 24 terms, then
`a_(1)+a_(24)=a_(5)+a_(20)=a_(10)+a_(15)`
Given `a_(1)+a_(5)+a_(10)+a_(15)+a_(20)+a_(24)=225`
or `(a_(1)+a(24))+(a_(5)+a_(20))+(a_(10)+a_(15))=225`
or `3(a_(1)+a_(24))=225`
or `a_(1)+a_(24)=225/3=75` (1)
`thereforeS_(24)=24/2(a_(1)+a_(24))` [Using `S_(n)=n/2(a_(1)+a_(n))`]
=12(75)
=900 [Using (1)]