If a,b,c are in GP and a + x , b + x, c + x are in HP. Find x where a,b,c are distinct numbers?

2 Answers

x=b.x=b.

Explanation:

As a,ba,b and cc are in GP, we have b^2=acb2=ac.

Further as a,b,ca,b,c are distinct numbers, a!=b!=c!=aabca

If a+xa+x, b+xb+x and c+xc+x are in HP, then

1/(a+x),1/(b+x)1a+x,1b+x and 1/(c+x)1c+x are in AP and therefore

1/(c+x)-1/(b+x)=1/(b+x)-1/(a+x)1c+x1b+x=1b+x1a+x

i.e. (b+x-c-x)/((c+x)(b+x))=(a+x-b-x)/((b+x)(a+x))b+xcx(c+x)(b+x)=a+xbx(b+x)(a+x)

i.e. (b-c)/(bc+bx+cx+x^2)=(a-b)/(ba+bx+ax+x^2)bcbc+bx+cx+x2=abba+bx+ax+x2

or (b-c)(ba+bx+ax+x^2)=(a-b)(bc+bx+cx+x^2)(bc)(ba+bx+ax+x2)=(ab)(bc+bx+cx+x2)

or b^2a+b^2x+abx+bx^2-abc-bcx-acx-cx^2=abc+abx+acx+ax^2-b^2c-b^2x-bcx-bx^2b2a+b2x+abx+bx2abcbcxacxcx2=abc+abx+acx+ax2b2cb2xbcxbx2

or b^2a+2b^2x+2bx^2-2abc-2acx-cx^2-ax^2+b^2c=0b2a+2b2x+2bx22abc2acxcx2ax2+b2c=0

or x^2(2b-c-a)+x(2b^2-2ac)+b^2a-2abc+b^2c=0x2(2bca)+x(2b22ac)+b2a2abc+b2c=0

As b^2=acb2=ac, this becomes x^2(2b-c-a)+b^2(a-2b+c)=0,x2(2bca)+b2(a2b+c)=0,

or x^2(2b-c-a)-b^2(2b-c-a)=0,x2(2bca)b2(2bca)=0,

rArr (2b-c-a)(x^2-b^2)=0.(2bca)(x2b2)=0.

rArr {(b-c)+(b-a)}}(x-b)(x+b)=0.{(bc)+(ba)}}(xb)(x+b)=0.

Since, anebnec, (b-c)ne0,(b-a)ne0, :.(b-c)+(c-a)ne0.

Also, x+bne0.

Clearly, x=b.

answer is +2 and not equal to -2

Explanation:

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