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.

Let the a be first term and d be common difference of AP

And we know that, that the nth term is

a_n = a + (n - 1)d

Given,

2a₈ = a₂

2(a + 7d) = a + d

2a + 14d = a + d

a = - 13d [ eqn1]

Also,

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

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

2a + 27d = 3

2 (- 13d) + 27d = 3 [ using eqn1]

d = 3

using this value in eqn1

a = - 13(3)

= - 39

Now,

a₁₅ = a + 14d

= - 39 + 14(3)

= - 39 + 42

= 3

So 15^th term is 3.