Question

Let $a_{1}, a_{2}, \ldots, a_{10}$ be in $A.P$. and $h_{1}, h_{2}, \ldots, h_{10}$ be in $H.P$. If $a_{1}=h_{1}=2 \& a_{10}=h_{10}=3$ then $a_{4} h_{7}$ is

Answer 1

$a_{1}, a_{2}, \ldots, a_{10}$, are in A.P.
$h_{1}, h_{2}, \ldots, h_{10}$ are in H.P.
$\therefore \frac{1}{h_{1}}, \frac{1}{h_{2}}, \ldots, \frac{1}{h_{10}}$ are in A.P.
$\therefore a_{10}=a_{1}+9 d$
$3=2+9 d $
$\therefore d=\frac{1}{9}$
and $\frac{1}{h_{10}}=\frac{1}{h_{1}}+9 D$;
$ \frac{1}{3}=\frac{1}{2}+9 D$;
$ -\frac{1}{6}=9 D ;$
$\therefore D=-\frac{1}{54}$
$\therefore a_{4}=a_{1}+3 d=2+\frac{1}{3}=\frac{7}{3} $
$\frac{1}{h_{7}}=\frac{1}{h_{1}}+6 D$
$ \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} $
$\Rightarrow \frac{1}{h_{7}}=\frac{7}{18}$
$\therefore h_{7}=\frac{18}{7}$
$\therefore a_{4} h_{7}=\frac{7}{3} \times \frac{18}{7}=6$