Question

if a, b,and c are in AP then prove that ab+bc/ca,bc+ac/ab,ab+ac/bc are also in AP​

Answer 1

Correct question:

$\sf{If \ a, \ b, \ c \ are \ in \ A.P. \ then \ prove \ that}$

$\sf{\dfrac{ab+ac}{bc}, \ \dfrac{ab+bc}{ac}, \ \dfrac{bc+ac}{ab} \ are \ also \ in \ A.P.}$

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Answer:

Given:

$\sf{\leadsto{a, \ b \ and \ c \ are \ in \ A.P.}}$

To prove:

$\sf{\dfrac{ab+ac}{bc}, \ \dfrac{ab+bc}{ac}, \ \dfrac{bc+ac}{ab} \ are \ also \ in \ A.P.}$

Proof:

$\sf{Note: \ (If \ terms \ are \ in \ A.P.)}$

$\boxed{\sf{2\times \ t_{2}=t_{1}+t_{3}}}$

$\sf{If \ a, \ b, \ c \ are \ in \ A.P., \ we \ can \ say}$

$\boxed{\sf{2b=a+c...(1)}}$

$\sf{\longmapsto{t_{1}=\dfrac{ab+ac}{bc}=\dfrac{a(b+c)}{bc}}}$

$\sf{Multiply \ numerator \ and \ denominator \ by \ a, \ we \ get}$

$\sf{t_{1}=\dfrac{a^{2}(b+c)}{abc}}$

$\sf{\longmapsto{t_{2}=\dfrac{ab+bc}{ac}=\dfrac{b(a+c)}{ac}}}$

$\sf{Multiply \ numerator \ and \ denominator \ by \ b, \ we \ get}$

$\sf{t_{2}=\dfrac{b^{2}(a+c)}{abc}}$

$\sf{\longmapsto{t_{3}=\dfrac{bc+ac}{ab}=\dfrac{c(a+b)}{ab}}}$

$\sf{Multiply \ numerator \ and \ denominator \ by \ c, \ we \ get}$

$\sf{t_{3}=\dfrac{c^{2}(a+b)}{abc}}$

$\sf{In \ t_{1}, \ t_{2} \ and \ t_{3} \ \dfrac{1}{abc} \ is}$

$\sf{common, \ hence \ we \ can \ eliminate \ it}$

$\sf{\therefore{t_{1}=a^{2}(b+c),}}$

$\sf{t_{2}=b^{2}(a+c),}$

$\sf{t_{3}=c^{2}(a+b)}$

$\sf{We \ can \ say, \ t_{1}, \ t_{2}, \ t_{3} \ are \ in \ A.P.}$

$\sf{if \ 2\times \ t_{2}=t_{1}+t_{3}}$

$\sf{R.H.S.=a^{2}(b+c)+c^{2}(a+b)}$

$\sf{=a^{2}b+a^{2}c+c^{2}a+c^{2}b}$

$\sf{=a^{2}c+c^{2}a+a^{2}b+c^{2}b}$

$\sf{=ac(a+c)+b(a^{2}+c^{2})}$

$\sf{=ac(a+c)+b[(a+c)(a+c)-2ac}$

$\sf{Substitute \ a+c=2b \ from \ equation(1), \ we \ get}$

$\sf{=2abc+b[2b(a+c)-2ac]}$

$\sf{=2abc+2b^{2}(a+c)-2abc}$

$\sf{=2b^{2}(a+c)...(2)}$

$\sf{L.H.S.=2[b^{2}(a+c)}$

$\sf{=2b^{2}(a+c)...(3)}$

$\sf{...from \ (2) \ and \ (3), \ we \ get}$

$\sf{L.H.S.=R.H.S.}$

$\sf{Hence, \ proved.}$

$\sf\purple{\tt{\dfrac{ab+ac}{bc}, \ \dfrac{ab+bc}{ac}, \ \dfrac{bc+ac}{ab} \ are \ also \ in \ A.P.}}$