Sequences and Series 3 Question 6

6. If a,b and c be three distinct real numbers in GP and a+b+c=xb, then x cannot be

(2019 Main, 9 Jan I)

(a) 4

(b) 2

(c) -2

(d) -3

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Answer:

Correct Answer: 6. (b)

Solution:

  1. Let b=ar and c=ar2, where r is the common ratio.

We know that, r+1r2( for r>0)

and r+1r2( for r<0)[ using AMGM]

1+r+1r3 or 1+r+1r1x3 or x1x(,1][3,)

Hence, x cannot be 2 .

Alternate Method

From Eq. (i), we have

1+r+r2=xrr2+(1x)r+1=0

For real solution of r,D0.

(1x)240x22x30(x3)(x+1)0++13

x(,1][3,)

x cannot be 2 .



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