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If a, b and c are three distinct real nu...

If a, b and c are three distinct real numbers in G.P. and a+b+c = xb, then x cannot be

A

4

B

2

C

`-2`

D

`-3`

Text Solution

Verified by Experts

The correct Answer is:
B
Let d = ar and `c = ar^(2)`, where r is the common ratio
Then, `a + b + c = xb`
`rArr a + ar + ar^(2) = xar`
`rArr 1 + r + r^(2) = x " ....(i) "[ :' a!=0]`
`rArr x = (1 + r + r^(2))/(r) = 1 + r + (1)/(r)`
We know that, `r + (1)/(r) ge 2` (for `r gt 0`)
and `r + (1)/(r) le -2` (for `r lt 0`) [using `AM ge GM`]
`:. 1 + r + (1)/(r) ge 3`
or `1 + r + (1)/(r) le -1`
`rArr x ge 3 or x le -1`
`rArr x in (-oo, -1] uu [3,oo)`
Hence, x cannot be 2
Alternate Method
From Eq. (i), we have
`1 + r + r^(2) = xr`
`rArr r^(2) + (1 -x)r + 1 = 0`
For real solution of `r, D ge 0`
`rArr (1 -x)^(2) - 4 ge 0`
`rArr x^(2) -2x - 3 ge 0`
`rArr (x -3) (x + 1) ge 0`

`rArr x in (-oo, -1] uu [3, oo)`
`:. x` cannot be 2.
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