If a,b,c are in AP then show that a(b+c)/bc,b(c+a)/ca and c(a+b)/ab
Answers
EXPLANATION.
If a, b and c are in ap.
As we know that,
Divide each term by abc, we get.
⇒ a/abc, b/abc, c/abc.
⇒ 1/bc, 1/ac, 1/ab. are also in ap.
Multiply each of the terms by ab + bc + ac in numerator, we get.
⇒ (ab + bc + ac)/bc, (ab + bc + ac)/ac, (ab + bc + ac)/ab.
⇒ [(ab + ac)/bc] + 1, [(ab + bc)/ac] + 1, [(bc + ac)/ab] + 1.
This terms are also in ap.
Subtract 1 from each terms of the equation, we get.
⇒ [(ab + ac)/bc] + 1 - 1, [(ab + bc)/ac] + 1 - 1, [(bc + ac)/ab] + 1 - 1.
⇒ (ab + ac)/bc, (ab + bc)/ac, (bc + ac)/ab.
⇒ a(b + c)/bc, b(a + c)/ac, c(a + b)/ab.
Hence Proved.
If a, b and c are in ap.
As we know that,
Divide each term by abc, we get.
⇒ a/abc, b/abc, c/abc.
⇒ 1/bc, 1/ac, 1/ab. are also in ap.
Multiply each of the terms by ab + bc + ac in numerator, we get.
⇒ (ab + bc + ac)/bc, (ab + bc + ac)/ac, (ab + bc + ac)/ab.
⇒ [(ab + ac)/bc] + 1, [(ab + bc)/ac] + 1, [(bc + ac)/ab] + 1.
This terms are also in ap.
Subtract 1 from each terms of the equation, we get.
⇒ [(ab + ac)/bc] + 1 - 1, [(ab + bc)/ac] + 1 - 1, [(bc + ac)/ab] + 1 - 1.
⇒ (ab + ac)/bc, (ab + bc)/ac, (bc + ac)/ab.
⇒ a(b + c)/bc, b(a + c)/ac, c(a + b)/ab.
Hence Proved.