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

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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 $a_{1}=h_{1}=2$ and $a_{10}=h_{10}=3$, then $a_{4} h_{7}$ is\

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(d) : Let $a_{2}, a_{3}, \ldots \ldots . ., a_{9}$ are 8 arithmetic means between $a_{1}=2$ and $a_{10}=3$.\ $	herefore a_{10}=10^{	ext {th }}$ term of an A.P. whose first term is $a_{1}=2$\ or $\quad 3=a_{1}+9 d=2+9 d \Rightarrow d=\frac{1}{9}$\ Now, $a_{4}=a_{1}+3 d=2+3\left(\frac{1}{9}ight)=2+\frac{1}{3}=\frac{7}{3}$\ Again let $h_{2}, h_{3}, \ldots \ldots, h_{9}$ are 8 harmonic means between $h_{1}=2, h_{10}=3$.\ i.e., $\frac{1}{h_{2}}, \frac{1}{h_{3}}, \ldots . . \frac{1}{h_{9}}$ are 8 arithmetic means between $\frac{1}{2}$ and $\frac{1}{3}$.\ $	herefore \frac{1}{3}=\frac{1}{2}+9 D \Rightarrow D=\frac{-1}{54}$\ $	herefore \frac{1}{h_{7}}=\frac{1}{h_{1}}+6 D=\frac{1}{2}-\frac{6}{54}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}$\ or $\quad h_{7}=\frac{18}{7} \quad 	herefore a_{4} h_{7}=\left(\frac{7}{3}ight)\left(\frac{18}{7}ight)=6$\