Question

Let $a_1,a_2,\cdots ,a_{10}$ be in A.P. and $h_1,h_2,\cdots ,h_{10}$ be in H.P. 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

The correct option is D 6

Let d be the common difference of the A.P.
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 $\frac{1}{h_1},\frac{1}{h_2},\cdots ,\frac{1}{h_{10}}$
Then, $h_{10}=3\Rightarrow \frac{1}{h_{10}}=\frac{1}{3}$
$\Rightarrow \frac{1}{h_1}+9D=\frac{1}{3}\Rightarrow \frac{1}{2}+9D=\frac{1}{3}$
$\Rightarrow D=-\frac{1}{54}$
$\therefore \frac{1}{h_7}=\frac{1}{h_1}+6D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}$
$\Rightarrow h_7=\frac{18}{7}$
$\therefore a_4{h}_7=\frac{7}{3}\times \frac{18}{7}=6$