JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q6

MathematicsSingle correct+4 / -1Question 6 of 90

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

\dfrac{35}{27}

\dfrac{27}{28}

\dfrac{28}{27}

Answer and worked solution

Official answer: D

First A.P. $d=\frac19$ , so $a_4=2+\frac39=\frac73$ .

Second A.P. $b_1=\frac12$ , $b_{10}=\frac13$ , so $d'=-\frac1{54}$ and $b_4=\frac12-\frac1{18}=\frac49$ .

Product. $\frac73\cdot\frac49=\frac{28}{27}$ .

(D)

Topics:Arithmetic progression

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 6