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If m is the A.M. of two distinct real numbers `l` and `n""(""l ,""n"">""1)` and G1, G2 and G3 are three geometric means between `l` and n, then `G1 ^4+2G2 ^4+G3^ 4` equals, (1) `4l^2` mn (2) `4l^m^2` mn (3) `4l m n^2` (4) `4l^2m^2n^2`

A

`4l^(2)m^(2)n^(2)`

B

`4l^(2)mn`

C

`4lm^(2)n`

D

`4lmn^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C

Given, `m=(l+n)/2rArr l+n =2n" ...(i)"`
and l, `G_1, G_2, G_3,` n are in GP
`:." "G_1/l =G_2/G_1=G_2/G_2=n/G_3`
`rArr" "G_1G_3 = "In," G_1^2 = lG_2, G_2^2 =G_3 G_1, G_3^2 =nG_21" (ii)"`
Now, `G_1^4 +G_2^4 +G_3^4 =l^2 G_2^2+2G_2^4+n^2G_2^2`
`=G_2^2(l^2 +2G_2^2 +n^2)" "["from eq.(ii)"]`
`=G_3 G_1 (l^2 +2G_2^2 +n^2) " " [from Eq.(ii) ]`
`=G_3 G_1 (l^2 +2G_3 G_1 + n^2)`
= In `(l^2+"2 In"+n^2)" " [from Eq.(ii)]`
= In `(l+n)^2` = In `(2m)^2" "[from Eq. (i)]`
`=4//m^2n`
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