Question

The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.

Answer 1

We know that,

First term of an AP = a

Common difference of AP = d

n^th term of an AP, a_n = a + (n – 1)d

According to the question,

a_s = ½ a₂

2a₈ = a₂

2(a + 7d) = a + d

2a + 14d = a + d

a = – 13d …(1)

Also,

a₁₁ = 1/3 a₄ + 1

3(a + 10d) = a + 3d + 3

3a + 30d = a + 3d + 3

2a + 27d = 3

Substituting a = -13d in the equation,

2 (- 13d) + 27d = 3

d = 3

Then,

a = – 13(3)= – 39

Now,

a₁₅  = a + 14d

= – 39 + 14(3)

= – 39 + 42

= 3

So 15^th term is 3.