Question

Let a₁, a₂, .... , a₁₀ be in A.P. and h₁, h₂, .... , h₁₀ be in H.P. If a₁ = h₁ = 2 and a₁₀ = h₁₀ = 3, then a₄h₇ is

(a) 2 

(b) 3 

(c) 5 

(d) 6

Answer 1

(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.