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Let a1,a2,a3 ….and b1 , b2 , b3 … be two...

Let `a_1,a_2,a_3` ….and `b_1 , b_2 , b_3 …` be two geometric progressions with `a_1= 2 sqrt(3)` and `b_1= 52/9 sqrt(3)` If `3a_99b_99=104` then find the value of `a_1 b_1+ a_2 b_2+…+a_nb_n`

Text Solution

Verified by Experts

The correct Answer is:
3536
`a_(1)=2sqrt3andb_(1)=52/9sqrt3`
Also, `3a_(99)b_(99)=104`
`rArra_(1)r_(1)^(98)b_(1)r_(2)^(98)=104//3`
`rArrr_(1)^(98)r_(2)^(98)=1` (putting the values of `a_(1)` and `b_(1)`)
`rArrr_(1)r_(2)=1`
`rArra_(i)b_(i)=104/3AAi`
`thereforea_(1)b_(1)+a_(2)b_(2)+...+a_(n)b_(n)=102xx104/3`
`=34xx104`
=3536
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