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 m is

Options

A. 2044
B. 1022
C. 511
D. None of these

Answer

D

Explanation

`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)"`
`G_(1)+G_(2)+G_(3)+".....+"G_(n)=r+r^(2)+r^(3)+".........."+r^(n)`
`=r+r^(2)+r^(3)+".........."+r^(10)=r((1-r^(10)))/(1-r)`
`[:.r=((l)/(a))^((l)/(n+1))=((2187)/(1))^((1)/(11))=3^((7)/(11))]`
`3^((7)/(11))((1-3^((70)/(11))))/((1-3^((7)/(11))))`.