Let a_1, a_2, dots, a_{10} be in A.P. and h_1 | vedclass

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(D) Given that $a_1, a_2, \dots, a_{10}$ is an $A.P.$ with $a_1 = 2$ and $a_{10} = 3$.The $n$-th term of an $A.P.$ is $a_n = a_1 + (n-1)d$.For $n=10$,

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$6$

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(D) Given that $a_1, a_2, \dots, a_{10}$ is an $A.P.$ with $a_1 = 2$ and $a_{10} = 3$.The $n$-th term of an $A.P.$ is $a_n = a_1 + (n-1)d$.For $n=10$,$3 = 2 + 9d$,which gives $d = \frac{1}{9}$.Thus,$a_4 = a_1 + 3d = 2 + 3(\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}$.Given that $h_1, h_2, \dots, h_{10}$ is an $H.P.$ with $h_1 = 2$ and $h_{10} = 3$.Then $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_{10}}$ is an $A.P.$ with first term $\frac{1}{2}$ and $10$-th term $\frac{1}{3}$.Let the common difference of this $A.P.$ be $D$.$\frac{1}{h_{10}} = \frac{1}{h_1} + 9D \implies \frac{1}{3} = \frac{1}{2} + 9D \implies 9D = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} \implies D = -\frac{1}{54}$.Then $\frac{1}{h_7} = \frac{1}{h_1} + 6D = \frac{1}{2} + 6(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{9} = \frac{9-2}{18} = \frac{7}{18}$.Therefore,$h_7 = \frac{18}{7}$.Finally,$a_4 h_7 = \frac{7}{3} 	imes \frac{18}{7} = 6$.

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 $a_4 h_7$ is

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