If a1,a2,a3,…,a24 are  in arithmetic progression  and  a1+a5+a10+a15+a20+a24=225 then a1+a2+a3+…+a23+a24 is equal to

Given that,  a1+a5+a10+a15+a20+a24=225⇒    a1+a24+a5+a20+a10+a15=225⇒    3a1+a24=225⇒    a1+a24=75[∵in an AP the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of the first and last term] ∴ a1+a2+…+a24=n2(a+l)=242a1+a24=12×75=900  [from eq.(i)]