Question

The 10th term of an AP is 29 and the sum of the first 20 term is 610.find the sum of the first 30 terms.

Answer 1

a+9d=29--------------------1equation
s(20)=20/2(2a+9d)=610
10(2a+9d)=610
20a+90d=610
20a=610-90d
a=(610-90d)/20----------------------------2equation
from equation 1
a=29-9d------------------------------------------3equation
from eq1and2
29-9d=(610-90d)/20
20(29-9d)=610-90d
580-180d=610-90d
-180d+90d=610-580
-90d=30
d=30/90=-1/3put in 3equation
a=29-9(-1/3)
a=29+3
a=32
now here
to find30th term sum
s(30)=n/2(2a+29d)
s(30)=15(2(32)+29(-1/3)
s(30)=15(64-29/3)
  15(192-29)/3
5(163)
s(30)=815