(d) 6
Let d be the common difference of the given A.P, a1, a2, ...., a10.
Then, given, a1 = 2 and a10 = a1 + 9d = 3
⇒ 2 + 9d = 3 ⇒ d = \(\frac{1}{9}\)
⇒ a4 = a1 + 3d = 2 + 3 × \(\frac{1}{9}\) = 2\(\frac{1}{3}\) = \(\frac{7}{3}\) .....(i)
Now, h1, h2, .... , h10 are in H.P.
⇒ \(\frac{1}{h_1}\), \(\frac{1}{h_2}\),......, \(\frac{1}{h_{10}}\) are in A.P.
Let d1 be the common difference of this A.P.
Given, h1 = 2 ⇒ First term of the A.P = \(\frac{1}{h_1}\) = \(\frac{1}{2}\)
Also, h10 = 3 ⇒ \(\frac{1}{h_{10}}\) = \(\frac{1}{h_1}\) + 9d1
⇒ \(\frac{1}{3}\) = \(\frac{1}{2}\) + 9d1 ⇒ 9d1 = \(\frac{1}{3}\) - \(\frac{1}{2}\) = \(-\frac{1}{6}\)
⇒ d1 = \(-\frac{1}{54}\)
∴ \(\frac{1}{h_7}\) = \(\frac{1}{h_1}\) + 6d1 = \(\frac{1}{2}\) + 6 x \(-\frac{1}{54}\) = \(\frac{1}{2}\) - \(\frac{1}{9}\) = \(\frac{7}{18}\)
⇒ h7 = \(\frac{18}{7}\) ......(ii)
∴ From (i) and (ii) a4 h7 = \(\frac{7}{3}\) x \(\frac{18}{7}\) = 6.