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 $a_1=h_1=2\&a_{10}=h_{10}=3$ then $a_4{h}_7$ is

Answer 1

Answer: 6

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