Answer : (b) 6
Let a1, a2, a3, ..... , an be the n arithmetic means between 3 and 17.
Then, 3, a1, a2, a3, ..... , an, 17 form an A.P.
Let d be the common difference of this A.P.
Then, a1 = 3 + d and an = 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.