Question

Let \( a_1, a_2, \dots, a_{40} \) be in AP and \( h_1, h_2, \dots, h_{10} \) be in HP. If \( a_1 = h_1 = 2 \) and \( a_{10} = h_{10} = 3 \), then \( a_4 h_7 \) is:

• A. 2
• B. 3
• C. 5
• D. 6

Hint:

In problems involving arithmetic and harmonic progressions, remember to first solve for the common difference (AP) or the reciprocal terms (HP) and then use these to find the required terms.

Answer 1

The Correct Option is D

Let \( d \) be the common difference of the AP. Then, \[ a_{10} = 3 \Rightarrow a_1 + 9d = 3 \] \[ \Rightarrow 2 + 9d = 3 \Rightarrow d = \frac{1}{9} \] \[ \therefore a_4 = a_1 + 3d = 2 + \frac{1}{3} = \frac{7}{3} \] Let \( D \) be the common difference of \(\frac{1}{h_1}, \frac{1}{h_2}, \ldots, \frac{1}{h_{10}}\). Then, \[ h_{10} = 3 \] \[ \Rightarrow \frac{1}{h_{10}} = \frac{1}{3} \Rightarrow \frac{1}{2} + 9D = \frac{1}{3} \] \[ \Rightarrow 9D = -\frac{1}{6} \Rightarrow D = -\frac{1}{54} \] \[ \therefore \frac{1}{h_7} = \frac{1}{h_1} + 6D = \frac{1}{2} + \frac{7}{18} \] \[ \Rightarrow h_7 = \frac{18}{7} \] \[ \therefore a_4 = h_7 = \frac{7}{3} \times \frac{18}{7} = 6 \]