Let $a_{1}, a_{2}, . . . . a_{10}$ be in AP, and $h_{1}, h_{2}, . . . ., h_{10}$ be in HP. If $a_{1} = h_{1} = 2$ and $a_{10} = h_{10} = 3$ , then $a_{4} h_{7}$ is?

2

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3

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5

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6

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Solution

The correct option is D $6$
$a_{10} = a_{1} + 9 d_{1} ⟹ d_{1} = \frac{3 - 2}{9} = \frac{1}{9}$
Now, $a_{4} = a_{1} + 3 d_{1} = 2 + 3 \times (\frac{1}{9}) = \frac{7}{3} ∴ (\frac{1}{h_{10}}) = (\frac{1}{h_{1}}) + 9 d_{2} ⟹ (\frac{1}{3}) = (\frac{1}{2}) + 9 d_{2} ⟹ d_{2} = (\frac{- 1}{54})$
Now, $(\frac{1}{h_{7}}) = (\frac{1}{h_{1}}) + 6 d_{2} = \frac{1}{2} + 6 (\frac{- 1}{54}) (\frac{1}{h_{7}}) = \frac{7}{18}$
So, $a_{4} h_{7} = \frac{7}{3} \times \frac{18}{7} = 6$

Hence, this is the answer.

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