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(D) Given $a_{1}, a_{2}, a_{3}, \ldots$ is an $A.P.$ with $a_{1}=2$ and $a_{10}=3$.Using $a_{n} = a_{1} + (n-1)d_{1}$,we have $3 = 2 + 9d_{1}$,so $d_{

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$\frac{28}{27}$

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(D) Given $a_{1}, a_{2}, a_{3}, \ldots$ is an $A.P.$ with $a_{1}=2$ and $a_{10}=3$.Using $a_{n} = a_{1} + (n-1)d_{1}$,we have $3 = 2 + 9d_{1}$,so $d_{1} = \frac{1}{9}$.Thus,$a_{4} = a_{1} + 3d_{1} = 2 + 3(\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}$.Given $b_{1}, b_{2}, b_{3}, \ldots$ is an $A.P.$ with $a_{1}b_{1}=1$ and $a_{10}b_{10}=1$.Since $a_{1}=2$,$b_{1} = \frac{1}{2}$. Since $a_{10}=3$,$b_{10} = \frac{1}{3}$.Using $b_{n} = b_{1} + (n-1)d_{2}$,we have $\frac{1}{3} = \frac{1}{2} + 9d_{2}$,so $9d_{2} = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$,which gives $d_{2} = -\frac{1}{54}$.Thus,$b_{4} = b_{1} + 3d_{2} = \frac{1}{2} + 3(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{18} = \frac{9-1}{18} = \frac{8}{18} = \frac{4}{9}$.Therefore,$a_{4}b_{4} = (\frac{7}{3})(\frac{4}{9}) = \frac{28}{27}$.

If $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1}b_{1}=1=a_{10}b_{10}$,then $a_{4}b_{4}$ is equal to

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