- Home
- Standard 11
- Mathematics
8. Sequences and Series
medium
જો $a _{1}, a _{2}, a _{3} \ldots$ અને $b _{1}, b _{2}, b _{3} \ldots$ એ સમાંતર શ્રેણી મા હોય તથા $a_{1}=2, a_{10}=3, a_{1} b_{1}=1=a_{10} b_{10}$ હોય,તો $a_{4} b_{4}=\dots$
A
$\frac{35}{27}$
B
$1$
C
$\frac{27}{28}$
D
$\frac{28}{27}$
(JEE MAIN-2022)
Solution
$a_{1}, a_{2}, a_{3} \ldots \text { A.P. } ; a_{1}=2 ; a_{10}=3 ; d_{1}=\frac{1}{9}$
$b _{1}, b _{2}, b _{3}, \ldots$ $A.P.$ $; b _{1}=\frac{1}{2} ; b _{10}=\frac{1}{3} ; d _{2}=\frac{-1}{54}$
[Using $a_{1} b_{1}=1=a_{10} b_{10} ; d_{1}$ and $d_{2}$ are common differences respectively]
$a _{4} \cdot b _{4} =\left(2+3 d _{1}\right)\left(\frac{1}{2}+3 d _{2}\right)$
$=\left(2+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{18}\right)$
$=\left(\frac{7}{3}\right)\left(\frac{8}{18}\right)=\frac{28}{27}$
Standard 11
Mathematics
