Question

Let $a_1,a_2,\dots .,a_{10}$ be in A.P. and $h_1,h_2,\dots ..,h_{10}$ be in H.P. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$ then $a_4{h}_7$ is equal to

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

Answer: (D)

$a_1,a_2,\dots .,a_{10}\text{, are in A.P. }$
$h_1,h_2,\dots ..,h_{10}\text{ are in H.P. }$
$\therefore \frac{1}{h_1},\frac{1}{h_2},\dots ..,\frac{1}{h_{10}}\text{ are in A.P. }$
$\therefore a_{10}=a_1+9d$
$3+2+9d$
$\therefore d=\frac{1}{9}$
$\text{ and }\frac{1}{h_{10}}=\frac{1}{h_1}+9D;\frac{1}{3}=\frac{1}{2}+9D;-\frac{1}{6}=9D;$
$\therefore D=-\frac{1}{54}$
$\therefore a_4=a_1+3d=2+\frac{1}{3}=\frac{7}{3}$
$\frac{1}{h_7}=\frac{1}{h_1}+6D\Rightarrow \frac{1}{h_7}=\frac{1}{2}+6\times \left(-\frac{1}{54}\right)$
$\Rightarrow \frac{1}{h_7}=\frac{1}{2}-\frac{1}{9}\Rightarrow \frac{1}{h_7}=\frac{7}{18}$
$\therefore h_7=\frac{18}{7};$
$\therefore a_4{h}_7=\frac{7}{3}\times \frac{18}{7}=6$