How many terms are there in the AP 7, 11, 15, …, 139 ?

1.	How many terms are there in the AP 7, 11, 15, …, 139 ?
Answer» Given AP is 7, 11, 15, …., 139. First term of AP is a = 7. And common difference of AP is d = a₂ – a₁ = 11 – 7 = 4. Let n^th term of AP is 139. ∴ a + (n – 1) d = 139. (∵ n^th term of AP is given by a_n = a + (n – 1) d) ⇒ 7 + (n – 1)4 = 139 (∵ a = 7 & d = 4) ⇒ 4(n – 1) = 139 – 7 = 132 ⇒ n – 1 = 132 4 = 33 ⇒ n = 33 + 1 = 34. Hence, total term in given AP is 34.

Answer»

Given AP is 7, 11, 15, …., 139.

First term of AP is a = 7.

And common difference of AP is d = a₂ – a₁ = 11 – 7 = 4.

Let n^th term of AP is 139.

∴ a + (n – 1) d = 139. (∵ n^th term of AP is given by a_n = a + (n – 1) d)

⇒ 7 + (n – 1)4 = 139 (∵ a = 7 & d = 4)

⇒ 4(n – 1) = 139 – 7 = 132

⇒ n – 1 = 132 4 = 33

⇒ n = 33 + 1 = 34.

Hence, total term in given AP is 34.

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How many terms are there in the AP 7, 11, 15, …, 139 ?

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