Question

The first term of an AP is 14 and the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign . what is the 3rd term of the AP???

Answer 1

first term =14
common diff=d
==> 5/2(2*14+(4)d)=-10/2(2*4+9d)
solve it 
d=-42/11
t3=a+ 2d=70/11

Answer 2

A1=14

s(5)=-s(10)

5/2(2*14+(5-1)d=-10(2*14+(10-1)d

5(28+4d)=-10(28+9d)

10(14+2d)=-10(28+9d)

d=-42/11

a3=a+2d

a3=14+2(-42/11)

154-84/11

70/11

6*4/11