1.

If a1,a2,a3,..........are in ap such that a1+a5+a10+a15+a20+a24=225, then a1+a2+a3+..........a23+a24 is

Answer»

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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