Let {a_1, a_2, dots, a_{10}} be in A.P. and { | vedclass

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(D) Given ${a_1 = h_1 = 2}$ and ${a_{10} = h_{10} = 3}$.For the $A.P.$,${a_{10} = a_1 + 9d = 3}$,so ${2 + 9d = 3}$,which gives ${d = \frac{1}{9}}$.Thu

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$6$

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(D) Given ${a_1 = h_1 = 2}$ and ${a_{10} = h_{10} = 3}$.For the $A.P.$,${a_{10} = a_1 + 9d = 3}$,so ${2 + 9d = 3}$,which gives ${d = \frac{1}{9}}$.Thus,${a_4 = a_1 + 3d = 2 + 3(\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}}$.For the $H.P.$,the reciprocals $\frac{1}{h_n}$ are in $A.P.$ Let ${H_n = \frac{1}{h_n}}$.Then ${H_1 = \frac{1}{2}}$ and ${H_{10} = \frac{1}{3}}$.${H_{10} = H_1 + 9D = \frac{1}{3}}$,so ${9D = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}}$,which gives ${D = -\frac{1}{54}}$.Then ${H_7 = H_1 + 6D = \frac{1}{2} + 6(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{9} = \frac{9-2}{18} = \frac{7}{18}}$.Since ${h_7 = \frac{1}{H_7}}$,we have ${h_7 = \frac{18}{7}}$.Therefore,${a_4 h_7 = \frac{7}{3} 	imes \frac{18}{7} = 6}$.

Let ${a_1, a_2, \dots, a_{10}}$ be in $A.P.$ and ${h_1, h_2, \dots, h_{10}}$ be in $H.P.$ If ${a_1 = h_1 = 2}$ and ${a_{10} = h_{10} = 3}$,then the value of ${a_4 h_7}$ is:

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