Question

Let $a_1,a_2,...,a_{10}$ be in AP and $h_1, h_2$ equal to .....,$h_{10}$ 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

Answer 1

Answer: (D)

6

Answer 2

Since $a_1,a_2,a_3...,a_{10}$ are in AP.
Now, \hspace15mm a_{10} =a_1 +9d
\Rightarrow\hspace25mm 3 = 2 +9d
\Rightarrow\hspace25mm d = 1/9
and \hspace20mm a_4 = a_1 + 3d
\Rightarrow\hspace20mm a_4 = a_1 + 3(1/9) = 2 +1/3 = 7/3
Also, $h_1, h_2, h_3,...,h_{10}$ are in HP.
\Rightarrow\hspace25mm \frac{1}{h_1}, \frac{1}{h_2},\frac{1}{h_3},...,\frac{1}{h_{10}} are in AP.
Given \hspace15mm h_1 = 2, h_{10} = 3
\therefore \hspace20mm \frac{1}{h_{10}}=\frac{1}{h_1} + 9d_1
\Rightarrow \hspace20mm \frac{1}{3}=\frac{1}{2} + 9d_1
\Rightarrow \hspace15mm -\frac{1}{6}=9d_1 \Rightarrow d_1 = - \frac{1}{54}
and \hspace15mm \frac{1}{h_7}=\frac{1}{h_1} + 6d_1
\Rightarrow \hspace20mm \frac{1}{h_7}= \frac{1}{2} + \frac{6 \times 1}{-54} \Rightarrow \frac{1}{h_7} = \frac{1}{2}- \frac{1}{9}
\Rightarrow \hspace20mm \frac{1}{h_7}= \frac{18}{7}
\therefore \hspace20mm a_4 h_7=\frac{7}{3} \times \frac{18}{7} = 6