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Let $a_1, a_2, \ldots \ldots a_{40}$ be in AP and $h_1, h_2, \ldots . h_{10}$ be in HP. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$, then $a_4 h_7$ is
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The correct answer is:
6
6
Let $d$ be the common difference of the AP. Then,
$\begin{aligned}
& a_{10}=3 \Rightarrow a_1+9 d=3 \\
\Rightarrow & 2+9 d=3 \Rightarrow d=\frac{1}{9} \\
\therefore & a_4=a_1+3 d=2+\frac{1}{3}=\frac{7}{3}
\end{aligned}$
Let $D$ be the common difference of $\frac{1}{h_1}, \frac{1}{h_2}, \ldots .-\frac{1}{h_{10}}$.
Then, $\quad h_{10}=3$
$\begin{array}{ll}
\Rightarrow & \frac{1}{h_{10}}=\frac{1}{3} \Rightarrow \frac{1}{2}+9 D=\frac{1}{3} \\
\Rightarrow & 9 D=-\frac{1}{6} \Rightarrow D=-\frac{1}{54} \\
\therefore & \frac{1}{h_7}=\frac{1}{h_1}+6 D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18} \\
\Rightarrow & h_7=\frac{18}{7} \\
\therefore & a_4 h_7=\frac{7}{3} \times \frac{18}{7}=6
\end{array}$
$\begin{aligned}
& a_{10}=3 \Rightarrow a_1+9 d=3 \\
\Rightarrow & 2+9 d=3 \Rightarrow d=\frac{1}{9} \\
\therefore & a_4=a_1+3 d=2+\frac{1}{3}=\frac{7}{3}
\end{aligned}$
Let $D$ be the common difference of $\frac{1}{h_1}, \frac{1}{h_2}, \ldots .-\frac{1}{h_{10}}$.
Then, $\quad h_{10}=3$
$\begin{array}{ll}
\Rightarrow & \frac{1}{h_{10}}=\frac{1}{3} \Rightarrow \frac{1}{2}+9 D=\frac{1}{3} \\
\Rightarrow & 9 D=-\frac{1}{6} \Rightarrow D=-\frac{1}{54} \\
\therefore & \frac{1}{h_7}=\frac{1}{h_1}+6 D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18} \\
\Rightarrow & h_7=\frac{18}{7} \\
\therefore & a_4 h_7=\frac{7}{3} \times \frac{18}{7}=6
\end{array}$
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