Question

If $a_1, a_2, a_3$ ….. and $b_1, b_2, b_3$ ….. are A.P., and $a_1 = 2, a_{10} = 3, a_1b_1 = 1 = a_{10}b_{10}$, then $a_4b_4$ is equal to

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

Answer 1

Answer: (D)

$\frac{28}{27}$

Answer 2

$a_1, a_2, a_3$ … are in A.P. (Let common difference is $d_1$)

$b_1, b_2, b_3$ … are in A.P. (Let common difference is $d_2$) and $a_1$ = $2$, $a_{10} = 3, a_1b_1 = 1 = a_{10}b_{10}$

∵ $a_1b_1 = 1$

$∴b1=\frac{1}{2}$

$a_{10}b_{10}= 1$

$∴b_{10}=\frac{1}{3}$

Now, $a_{10}=a_1+9d_1$

$⇒d_1=\frac{1}{9}$

$b_{10}=b_1+9d_2$

$⇒d_2=\frac{1}{9}[\frac{1}{3}−\frac{1}{2}]$

=$−\frac{1}{54}$

Now, $a_4=2+\frac{3}{9}=\frac{7}{3}$

$b_4=\frac{1}{2}–\frac{3}{54}=\frac{4}{9}$

$∴ \;a_4b_4=\frac{28}{27}$