If a1, a2, a3, .... is an A.P. such that a1+ a5+ a10+a15+a20+a24=225 , then a1+a2+a3+....+a23+a24  is equal to ‘N’ then sum of the given digits in N is

Since a1,a2,.........a24 are in A.P ,   a1+a24=a2+a23= a3+a22=..........=a12+a13       therefore a1+a24=a5+a20=a10+a15Hence the given relations reduce to, 3(a1+a24)=225 , giving a1+a24=75Hence S24=n2(a+l)=(24/2)(a1+a24)=12×75=900 =N sum of the digits of N = 9