The sum of the first seven terms of an AP is 182. If its 4^th and the 17^th terms are in the ratio 1: 5, find the AP.

Given sum of first seven terms of AP, S₇ = 182.

We know that sum of first n terms of AP, S_n = (n/2) [2a + (n – 1) d]

⇒ 182 = (7/2) [2a + (7 – 1) d]

⇒ 182 = (7/2) [2a + 6d]

⇒ 182 = (7/2) (2) [a + 3d]

⇒ 182/7 = a + 3d

⇒ a + 3d = 26 … (1)

Also given the 4^th and 17^th terms are in the ratio 1: 5.

We know that nth term, a_n = a + (n – 1) d

⇒ a₄ = a + (4 – 1) d = a + 3d

⇒ a₁₇ = a + (17 – 1) d = a + 16d

⇒ (a + 3d)/ (a + 16d) = 1/5

⇒ 5 (a + 3d) = 1 (a + 16d)

⇒ 5a + 15d = a + 16d

⇒ 4a = d … (2)

Substituting the above in (1),

⇒ a + 3 (4a) = 26

⇒ a + 12a = 26

⇒ 13a = 26

⇒ a = 2

Substituting a = 2 in (2),

⇒ d = 8

∴ The AP will be as follows: 2, 10, 18, 26, …