n arithmetic means are inserted between 3 and 17. If the ratio of the

1.	n arithmetic means are inserted between 3 and 17. If the ratio of the last and the first arithmetic mean is 3: 1, then n is equal to(a) 5 (b) 6 (c) 7 (d) 9
Answer» Answer : (b) 6 Let a₁, a₂, a₃, ..... , a_n be the n arithmetic means between 3 and 17. Then, 3, a₁, a₂, a₃, ..... , a_n, 17 form an A.P. Let d be the common difference of this A.P. Then, a₁ = 3 + d and a_n = 17 – d Given, \(\frac{a_n}{a_1} \) = \(\frac{3}{1}\) ⇒ \(\frac{17-d}{3+d}\) = \(\frac{3}{1}\) ⇒ 17 – d = 9 + 3d ⇒ 4d = 8 ⇒ d = 2. Also 17 is the (n + 2)th term of the given A.P. ∴ 17 = 3 + (n + 2 – 1)2 ⇒ 17 = 3 + (n + 1)2 ⇒ 14 = (n + 1)2 ⇒ n + 1 = 7 ⇒ n = 6.

n arithmetic means are inserted between 3 and 17. If the ratio of the last and the first arithmetic mean is 3: 1, then n is equal to(a) 5 (b) 6 (c) 7 (d) 9

Answer»

Answer : (b) 6

Let a₁, a₂, a₃, ..... , a_n be the n arithmetic means between 3 and 17.

Then, 3, a₁, a₂, a₃, ..... , a_n, 17 form an A.P.

Let d be the common difference of this A.P.

Then, a₁ = 3 + d and a_n = 17 – d

Given,

\(\frac{a_n}{a_1} \) = \(\frac{3}{1}\) ⇒ \(\frac{17-d}{3+d}\) = \(\frac{3}{1}\)

⇒ 17 – d = 9 + 3d

⇒ 4d = 8 ⇒ d = 2.

Also 17 is the (n + 2)th term of the given A.P.

∴ 17 = 3 + (n + 2 – 1)2

⇒ 17 = 3 + (n + 1)2

⇒ 14 = (n + 1)2

⇒ n + 1 = 7

⇒ n = 6.