Question

Let $a_1,a_2,....,a_{10}$ be in $A.P.$ and $h_1,h_2,....,h_{10}$ 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 $6$

Given that $a_1{,a}_2{,a}_3,...,a_{10}$ are in A.P.
Let $a_{10}=a_1+9d$
$\Rightarrow 3=2+9d$
$\Rightarrow d=\frac{1}{9}$
So, $a_4=a_1+3d$
$\Rightarrow a_4=2+3\left(\frac{1}{9}\right)$
$\Rightarrow a_4=\frac{7}{3}$
Now, given that $h_1,h_2{,...,h}_{10}$ are in H.P
$\Rightarrow \frac{1}{h_1},\frac{1}{h_2},...,\frac{1}{h_{10}}$ are in A.P
Let, $\frac{1}{h_{10}}=\frac{1}{h_1}+9d$
$\Rightarrow \frac{1}{3}=\frac{1}{2}+9d$
$\Rightarrow \frac{1}{3}-\frac{1}{2}=9d$
$\Rightarrow 9d=-\frac{1}{6}$
$\Rightarrow d=-\frac{1}{54}$
So, $\frac{1}{h_7}=\frac{1}{h_1}+6d$
$\Rightarrow \frac{1}{h_7}=\frac{1}{2}+6(-\frac{1}{54})$
$\Rightarrow \frac{1}{h_7}=\frac{1}{2}-\frac{1}{9}$
$\Rightarrow \frac{1}{h_7}=\frac{7}{18}$
$\Rightarrow h_7=\frac{18}{7}$
Consider, $a_4{h}_7=\frac{7}{3}\cdot \frac{18}{7}$
$\Rightarrow a_4{h}_7=6$