The sum of first n terms of an AP is given by S_n = 2n² + 3n. Find the sixteenth term of the AP.

Given: Sum of first n terms, S_n = 2n² + 3n …(i)

We need to find first term & common difference of this AP in order to find the sixteenth term.

So substituting n = 1 in equation (i), we get

S₁ = 2(1)² + 3(1)

⇒ S₁ = 2 + 3 = 5

⇒ a₁ = 5 [∵ Sum of first term is itself the first term, i.e., S₁ = a₁]

Again, substituting n = 2 in equation (i), we get

S₂ = 2(2)² + 3(2)

⇒ S₂ = 8 + 6 = 14

⇒ a₁ + a₂ = 14 [∵ S₂ = a₁ + a₂]

⇒ 5 + a₂ = 14 [∵ a₁ = 5]

⇒ a₂ = 14 – 5 = 9

If we know a₁ = 5 and a₂ = 9, we can calculate common D.

D = a₂ – a₁

⇒ D = 9 – 5 = 4

Now we can easily calculate sixteenth term of the AP, which is given by

a₁₆ = a₁ + (16 – 1)D [∵ a_n = a + (n – 1)D; n = 16]

⇒ a₁₆ = 5 + 15×4

⇒ a₁₆ = 5 + 60 = 65

Hence, sixteenth term of the AP is 65.