Show that the list of numbers 9, 12, 15, 18, … form an A.P. Find its 16th term and the nth term.

Here,
a₂ - a₁ = 12 - 9 = 3,
a₃ - a₂ = 15 - 12 = 3.

i.e. any term - preceding term = 3 = fixed number,

Hence, given list of numbers forms an A.P. with first term = a = 9 and common difference = d = 3.

We know that,

a_n = a + (n - 1)d = 9 + (n - 1) × 3 = 9 + 3n - 3 = 6 + 3n.

Since, a_n = 6 + 3n, so

a₁₆ = 6 + 3 × 16 = 6 + 48 = 54.

Hence, the a_n = 3n + 6 and a₁₆ = 54.