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Question

Let a,b,c be three distinct numbers, such that a+b+c=xS and a2+b2+c2=S2. If a,b,c are in G.P., then which of the following is/are correct ?

A
13<x2<3
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B
13<x2<1
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C
1<x2<3
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D
13x23
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Solution

The correct option is C 1<x2<3
Let b=ar, c=ar2, r±1 as numbers are distinct
a2+a2r2+a2r4=S2 ...(1)
a+ar+ar2=xS ...(2)
Squaring both sides
a2(1+r+r2)2=x2S2 ...(3)

Equation (1) divided by equation (3)
a2(1+r2+r4)a2(1+r+r2)2=S2x2S2
(1+r2)2r2(1+r+r2)2=1x2
1+r2r21+r+r2=1x2
x2=r+1r+1r+1r1

Putting
y=r+1r;|y|>2
x2=y+1y1
x2=1+2y1
Now, when
y>2y1>10<2y1<21<1+2y1<31<x2<3

When,
y<2y1<30>2y1>231>1+2y1>131>x2>13
x2(13,1)(1,3)


Alternate:
y=x2+1x21 (where x21) ...(4)
r±1, |y|=r+1r>2
x2+1x21>2
|x2+1|>2|x21|
(x2+1)2[2(x21)]2>0
{(x2+1)+[2(x21)]}{(x2+1)[2(x21)]}>0
(3x21)(3x2)>0
13<x2<3
From eqn(4), x21
x2(13,1)(1,3)

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