Let A1,A2,A3,……A11 be 11 Arithmetic means and H1,H2,H3,….H11 be 11 Harmonic means and G1,G2,G3,……G11 be 11 Geometric means between two numbers 1 and 9 then value of ∏k=15 Ak⋅G12−2k⋅H12−k is

Let A1H11=A2H10=A3H9=A4H8=A5H7=9G2G10=G4G8=G62=9∏k=15 Ak⋅G12−2k⋅H12−k=A1H11A2H10A3H9A4H8A5H7G10G8G6G4G2=95(9)2⋅3=315

Let A1,A2,A3,……A11 be 11 Arithmetic means and H1,H2,H3,….H11 be 11 Harmonic means and G1,G2,G3,……G11 be 11 Geometric means between two numbers 1 and 9 then value of ∏k=15 Ak⋅G12−2k⋅H12−k is

Let A1H11=A2H10=A3H9=A4H8=A5H7=9G2G10=G4G8=G62=9∏k=15 Ak⋅G12−2k⋅H12−k=A1H11A2H10A3H9A4H8A5H7G10G8G6G4G2=95(9)2⋅3=315

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$Let A_{1}, A_{2}, A_{3}, \dots \dots A_{11} be 11 Arithmetic means and H_{1}, H_{2}, H_{3}, \dots . H_{11} be 11 Harmonic means and G_{1}, G_{2}, G_{3}, \dots \dots G_{11} be 11 Geometric means$
$between two numbers 1 and 9 then value of \prod_{k = 1}^{5} (A_{k} \cdot G_{12 - 2 k} \cdot H_{12 - k}) is$

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$3^{5}$

$3^{10}$

$3^{11}$

$3^{15}$

answer is C.

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Detailed Solution

$\begin{array}{l} Let A_{1} H_{11} = A_{2} H_{10} = A_{3} H_{9} = A_{4} H_{8} = A_{5} H_{7} = 9 \\ G_{2} G_{10} = G_{4} G_{8} = {(G_{6})}^{2} = 9 \\ \prod_{k = 1}^{5} (A_{k} \cdot G_{12 - 2 k} \cdot H_{12 - k}) \\ = (A_{1} H_{11}) (A_{2} H_{10}) (A_{3} H_{9}) (A_{4} H_{8}) (A_{5} H_{7}) (G_{10} G_{8} G_{6} G_{4} G_{2}) \\ = (9^{5}) (9)^{2} \cdot 3 \\ = 3^{15} \end{array}$