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Question

Suppose ${d_1}$ is the common difference of the $A.P.$

${x_1},{x_2},....{x_n}$ then

$\because $ ${x_8} - {x_3} = 5{d_1} = 12 \Rightarr

Vedclass · Accepted Answer

$2560$

Vedclass · Answer

Suppose ${d_1}$ is the common difference of the $A.P.$

${x_1},{x_2},....{x_n}$ then

$\because $ ${x_8} - {x_3} = 5{d_1} = 12 \Rightarrow {d_1} = \frac{{12}}{5} = 2.4$

$ \Rightarrow {x_5} = {x_3} + 2{d_1} = 8 + 2 	imes \frac{{12}}{5} = 12.8$

Suppose ${d_2}$ is the common difference of the $A.P.$ $\frac{1}{{{h_1}}},\frac{1}{{{h_2}}},.....\frac{1}{{{h_n}}}$ then

$5{d_2} = \frac{1}{{20}} - \frac{1}{8} = \frac{{ - 3}}{{40}} \Rightarrow {d_2} = \frac{{ - 3}}{{200}}$

$\because$ $\frac{1}{{{h_{10}}}} = \frac{1}{{{h_7}}} + 3{d_2} = \frac{1}{{200}} \Rightarrow {h_{10}} = 200$

$ \Rightarrow {x_5}.{h_{10}} = 12.8 	imes 200 = 2560$

If $x_1 , x_2 , ..... , x_n$ and $\frac{1}{{{h_1}}},\frac{1}{{{h^2}}},......\frac{1}{{{h_n}}}$ are two $A.P' s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$, then $x_5. h_{10}$ equals

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