Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Given, a₁₁ = 38 and a₁₆ = 73

We know that a_n = a + (n – 1)d

Hence, a₁₁ = a + 10d = 38

And, a₁₆ = a + 15d = 73

Subtracting 11th term from 16th term, we get following:

a + 15d – a – 10d = 73 – 38

Or, 5d = 35

Or, d = 7

Substituting the value of d in 11th term we get;

a + 10 x 7 = 38

Or, a + 70 = 38

Or, a = 38 – 70 = - 32

Now 31st term can be calculated as follows:

a₃₁ = a + 30d

= - 32 + 30 x 7

= - 32 + 210 = 178