$\begin{array}{1 1} 2 \\ 3 \\ 5 \\ 6 \end{array}$

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Answer : (d) 6

Explanation : Let the AP = $a_{1},a_{2},a_{3}--------a_{10}$

$HP\;=h_{1},h_{2},h_{3}--------h_{10}$

$a_{1}=h_{1}=2=a$

$a_{10}=h_{10}=3$

$a_{10}=a+9d=2+9d=3$

$d=\frac{1}{9}$

$h_{10}=3\quad\;\frac{1}{3}=\frac{1}{2}+9D$

$D=\frac{-1}{54}$

$a_{4}=a+3d=2+\frac{3}{9}=\frac{7}{3}$

$h_{7}\qquad\;\frac{1}{h_{1}}=\frac{1}{h_{1}}+6D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}$

$h_{7}=\frac{18}{7}$

$a_{4}\;h_{7}=\frac{7}{3}*\frac{18}{7}=6.$

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