Let $a_{1}, a_{2}, . . . . . . . a_{10}$ be in A.P and $h_{1}, h_{2}, . . . . . . . h_{10}$ be in H.P. $a_{1} = h_{1} = 2$ 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 C $6$
Let $d_{1}$ be the common difference of A.P
and $d_{2}$ be the common difference of the sequence of A.P, which is obtained after reciprocated the terms
Then $a_{10} = a_{1} + 9 d_{1} = 2 + 9 d_{1} = 3$ (given)
Also
$h_{10} = \frac{1}{\frac{1}{2} + 9 d_{2}} = 3$ (given)
Solving for both $d_{1}$ and $d_{2}$ , we get

$d_{1} = \frac{1}{9}$ and $d_{2} = - \frac{1}{54}$
Therefore $a_{4} h_{7} = [2 + 3 d_{1}] \times ⎡ ⎢ ⎢ ⎣ \frac{1}{\frac{1}{2} + 6 d_{2}} ⎤ ⎥ ⎥ ⎦$ $= 6$
Hence, option 'D' is correct.

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