Sum of first 24 terms of the AP a1, a2, a3............, if it is known that a1 + a5 + a10 + a15 + a20 + a24 = 225 is:?

a1 + a5 + a10 +a15 + a20 + a24  = 225        (given)

(a1+a24) + (a5+a20) + (a10+a15) = 225            .................1

let first term is a & common difference is d then

 a1 = a       ,  a24 = a+23d   ,       a1+a24 = 2a+23d            ..........2

 

 a5 = a+4d ,  a20 = a+19d ,         a5+a20 = 2a+23d           ..............3

 

 a10 = a+9d , a15 = a+14d ,        a10+a15 = 2a+23d          ..............4

putting 2 , 3 , 4 in eq 1 we get

3(2a+23d) = 225

 2a+23d = 75             ...........5

now , a1 + a2 + a3 ............a24 = S24

   S24 = 24/2[2a+(24-10d)]

          =12(2a+23d)             ...............6

from 5 & 6

S24 = 12*75 = 900

approve if u like it

a1+a5+a10+a15+a20+a24=225 (given)a+a+4d+a+9d+a+14d+a+19d+a+23d=2256a+69d=2253(2a+23d)=3*752a+23d=752a=75-23dNow from Sn=(n/2)(2a+(n-1)d)......(1)n=24, 2a=75-23dPutting these values in equation (1)Sn=(24/2)(75-23d+(24-1)dSn=(12)(75-23d+23d)Sn=(12)75Sn=900