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Let A(1),A(2),A(3),"......."A(m) be arit...

Let `A_(1),A_(2),A_(3),"......."A_(m)` be arithmetic means between `-3` and 828 and `G_(1),G_(2),G_(3),"......."G_(n)` be geometric means between 1 and 2187. Product of geometric means is `3^(35)` and sum of arithmetic means is 14025.
The value of n is

A

45

B

30

C

25

D

10

Text Solution

Verified by Experts

The correct Answer is:
D
`A_(1),A_(2),A_(3),"........",A_(m)` are arithmetic means between `-3` and 828.
So, `A_(1)+A_(2)+"........"+A_(m)=m((a+b))/(2)`
`implies A_(1)+A_(2)+"........"+A_(m)=m((-3+288)/(2))`
`implies 14025=m((825)/(2))`
`" " " " [" given that sum of AM's=14025 "]`
`implies m=17xx2`
`:.m=34" " ".......(i)"`
Now, `G_(1),G_(2),G_(3),"........".G_(n)` be the GM's between 1 and 2187.
`:.G_(1)G_(2)G_(3)"....."G_(n)=(ab)^((n)/(2))`
`implies 3^(35)=(1xx2187)^((n)/(2)) " " implies 3^(35)=3^((7n)/(2))`
So, `35=(7n)/(2)`
`implies n=10" " "..........(ii)"`
`n=10" " " " [" by Eq.(ii) "]`.
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