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

Options

A. `xlt-1or,xgt3`
B. `xlt-3or,xgt2`
C. `xlt-4or,xgt3`
D. none of these

Answer

A

Explanation

We have,
`b^(2)=acanda+b+c=xb`
`rArr" "b^(2)=acandb(x-1)=a+c`
`rArr" "b^(2)=acandb^(2)(x-1)^(2)=(a+c)^(2)`
`rArr" "ac(x-1)^(2)=(a+c)^(2)`
`rArr" "((c)/(1)+1)^(2)-(x-1)^(2)(c)/(a)`
`rArr" "(c^(2))/(a^(2))+(c)/(a){2-(x-1)^(2)}+1=0`
`rArr" "{2-(x-1)^(2)}^(2)-4gt0" "[because(c)/(a)" is real and "c!=a]`
`rArr" "{2-(x-1)^(2)-2}{2-(x-1)^(2)+2}gt0`
`rArr" "(x-1)^(2)(4-x^(2)+2x-1)lt0`
`rArr" "x^(2)-2x-3gt0rArr(x-3)*x+1)gt0rArrxlt-1or,xgt3`