Given that an=3n−1.
The common difference of the AP, d=a2−a1
=3(2)−1−(3(1)−1)
=5−2=3.
We know that the nth term of an AP with first term a and common difference d is given by
an=a+(n−1)d.
⟹a3=a+2d=2+2×3=8,
a7=a+6d=2+6×3=20,
a13=a+12d=2+12×3=38 and
a8=a+7d=2+7×3=23.
a2=a+d=2+3=5
a6=a+5d=2+5×3=17
∴a3+a7=8+20=28, a2+a6=5+17=22 and a13−a8=38−23=15