Question

if a1 a2 a3.... are in ap such that a1+a5+a10+a15+a20+a25 =225 then find value of a1 +a2+a3+.......+a23+a24

Answer 1

the answer is 900. check in the image.

Answer 2

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