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

Answer:

#x=b.#

Explanation:

As #a,b# and #c# are in GP, we have #b^2=ac#.

Further as #a,b,c# are distinct numbers, #a!=b!=c!=a#

If #a+x#, #b+x# and #c+x# are in HP, then

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

#1/(c+x)-1/(b+x)=1/(b+x)-1/(a+x)#

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

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

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

or #b^2a+b^2x+abx+bx^2-abc-bcx-acx-cx^2=abc+abx+acx+ax^2-b^2c-b^2x-bcx-bx^2#

or #b^2a+2b^2x+2bx^2-2abc-2acx-cx^2-ax^2+b^2c=0#

or #x^2(2b-c-a)+x(2b^2-2ac)+b^2a-2abc+b^2c=0#

As #b^2=ac#, this becomes #x^2(2b-c-a)+b^2(a-2b+c)=0,#

or #x^2(2b-c-a)-b^2(2b-c-a)=0,#

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

# rArr {(b-c)+(b-a)}}(x-b)(x+b)=0.#

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

Also, #x+bne0.#

Clearly, #x=b.#

Answer:

answer is +2 and not equal to -2

Explanation:

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