The 7th term of an AP is –1 and its 16th term is 17. The nth term of the AP is

Foundation

Mathematics Foundation

nth term of AP

Question

(3n + 8)

(4n – 7)

(15 – 2n)

(2n – 15)

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T₇ = – 1 ⇒ a + 6d = – 1. ....(i)

T₁₆ = 17 ⇒ a + 15d = 17. ....(ii)

On Solving (i) and (ii), we get a = – 13 and d = 2.

$∴$ T_n = a + (n – 1)d = – 13 + (n – 1) × 2 = (2n – 15).