Five numbers are in H.P. The middle term is 1 and the ratio of the sec

1.	Five numbers are in H.P. The middle term is 1 and the ratio of the second and fourth terms is 2 : 1. Then, the sum of the first three terms is(a) \(\frac{11}{2}\) (b) 5 (c) 2 (d) \(\frac{14}{2}\)
Answer» (d) \(\frac{11}{2}.\) Let the terms of the corresponding A.P. be a – 2d, a – d, a, a + d, a + 2d. Given, a = 1 and \(\frac{a-d}{a+d}\) = \(\frac{1}{2}\) ⇒ 2a – 2d = a + d ⇒ a = 3d ⇒ d = \(\frac{1}{3}\)a = \(\frac{1}{3}\) (∵ a = 1) ∴ First three terms of A.P are 1 - \(\frac{2}{3}\), 1 - \(\frac{1}{3}\), 1, i.e., \(\frac{1}{3}\),\(\frac{2}{3}\),1 ⇒ First three terms of corresponding H.P are 3, \(\frac{3}{2}\),1 ∴ Required sum = 3 + \(\frac{3}{2}\) + 1 = 5\(\frac{1}{2}\) = \(\frac{11}{2}.\)

Five numbers are in H.P. The middle term is 1 and the ratio of the second and fourth terms is 2 : 1. Then, the sum of the first three terms is(a) \(\frac{11}{2}\) (b) 5 (c) 2 (d) \(\frac{14}{2}\)

Answer»

(d) \(\frac{11}{2}.\)

Let the terms of the corresponding A.P. be

a – 2d, a – d, a, a + d, a + 2d.

Given, a = 1 and \(\frac{a-d}{a+d}\) = \(\frac{1}{2}\)

⇒ 2a – 2d = a + d

⇒ a = 3d ⇒ d = \(\frac{1}{3}\)a = \(\frac{1}{3}\) (∵ a = 1)

∴ First three terms of A.P are 1 - \(\frac{2}{3}\), 1 - \(\frac{1}{3}\), 1, i.e., \(\frac{1}{3}\),\(\frac{2}{3}\),1

⇒ First three terms of corresponding H.P are 3, \(\frac{3}{2}\),1

∴ Required sum = 3 + \(\frac{3}{2}\) + 1 = 5\(\frac{1}{2}\) = \(\frac{11}{2}.\)