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$Let a_{1}, a_{3}, \dots . a_{10} be in A.P, and h_{1}, h_{2}, \dots h_{10} be in H.P. If a_{1} = h_{1} = 2 a$ $a_{10} = h_{10} = 3, then a_{4} h_{7} is$

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Correct option is D

Let 'd' is common difference of A.P ∴3=a10=2+9d⇒d=19, a4=2+3d=73 Let 'D' is common difference of 1h1,1h2,…⋅1h10 ∴13=1h10=12+9D⇒D=−1541h7=12+6−154=718, a4h7=73×187=6

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Let a1,a3,….a10 be in A.P, and h1,h2,…h10 be in H.P. If a1=h1=2 a a10=h10=3, then a4h7 is

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$Let a_{1}, a_{3}, \dots . a_{10} be in A.P, and h_{1}, h_{2}, \dots h_{10} be in H.P. If a_{1} = h_{1} = 2 a$ $a_{10} = h_{10} = 3, then a_{4} h_{7} is$