Question

If $a_1,a_2,a_3\dots$ and $b_1,b_2,b_3\dots .$ 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

• A. $\frac{35}{27}$
• B. $1$
• C. $\frac{27}{28}$
• D. $\frac{28}{27}$

Answer 1

Answer: (D)

$a_1,a_2,a_3\dots .A.P.;a_1=2;a_{10}=3;d_1=\frac{1}{9}$
$b_1,b_2,b_3,\dots$ A.P. $;b_1=\frac{1}{2};b_{10}=\frac{1}{3};d_2=\frac{-1}{54}$
[Using $a_1{b}_1=1=a_{10}{b}_{10};d_1\&d_2$ are common differences respectively]
$a_4\cdot b_4=\left(2+3d_1\right)\left(\frac{1}{2}+3d_2\right)$
$=\left(2+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{18}\right)$
$=\left(\frac{7}{3}\right)\left(\frac{8}{18}\right)=\frac{28}{27}$