Question

First term of an arithmetic progression is 2 if the sum of its first five terms is equal to one - fourth of the sum of the next five term then the sum of its first 30 terms is

Answer 1

a=2
s5=1/4(s10-s5)
s30= ?
sn=n/2(2a+(n-1)d
s5=5/2(2(2)+(5-1)d
s5=5/2(4+4d)
S10=5(4+9d)
s5=1/4(s10-s5)
5/2(4+4d)=1/4(5(4+9d)-5/2(4+4d)
10/2+20d/2=1/4(20+45d)-20/2-20d/2
5+10d=20/4+45d/4-10-10d
5+10+10d+10d-5=45d/4
15+20d-5=45d/4
20d+10=45d/4
10=45d/4-20d
10=45d-80d/4
10=-35d/4
10(4)=-35d
40/-35=d
8/-7=d

s30=30/2(2(2)+(30-1)8/-7)
s30=15(4)+29(8/-7))
=60+232/-7
=60-33.14
s30=27.86