Question

A1,a2,a3..... are in ap given that 4th term and7th term are in the ratio 3:2 then prove that a13 in that ap is 0

Answer 1

given a4:a7 = 3:2
i.e. 2×a4 = 3×a7
2×(a+3d) = 3×(a+6d)
2a+6d = 3a+18d
Therefore a+12d = 0
Thus 13th term = 0

Answer 2

a4 :a7 = 3:2
a + 3d / a + 6d = 3 / 2
2a + 6d = 3a + 18d
6d - 18d = 3a - 2a
=> -12d = a

a13 = a + 12d
As a = -12d
a13 = -12d + 12d
a13 = 0

Hence proved...