Last Activity: 14 Years ago
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
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Last Activity: 8 Years ago
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
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