If a_{1},a_{2},a_{3} and b_{1},b_{2},b_{3} are in 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: Sequence & Series (Mathematics)

(d): We have given, $a_1=2, a_{10}=3, a_1 b_1=1=a_{10} b_{10}$ Let $d_1$ be the common difference of an A.P. $a_1, a_2, a_3, \ldots \ldots$ and $d_2$ be the common difference of an A.P. $b_1, b_2, b_3 \ldots \ldots$ $\begin{aligned} & a_{10}=a_1+9 d_1 \Rightarrow 3=2+9 d_1 \Rightarrow d_1=\frac{1}{9} \\ & b_{10}=b_1+9 d_2 \Rightarrow \frac{1}{3}=\frac{1}{2}+9 d_2 \Rightarrow d_2=-\frac{1}{54} \\ & a_4=a_1+3 d_1=2+3\left(\frac{1}{9}\right)=\frac{7}{3} \\ & b_4=b_1+3 d_2=\frac{1}{2}+3\left(\frac{-1}{54}\right)=\frac{8}{18} \therefore a_4 b_4=\frac{28}{27}\end{aligned}$

Explanation

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