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if a^2(b+c), b^2(c+a), c^2(a+b) are in A.P, prove that either a,b,c are in A.P or ab+bc+ca=0

if a^2(b+c), b^2(c+a), c^2(a+b) are in A.P, prove that either a,b,c are in A.P or ab+bc+ca=0

Grade:11

2 Answers

Vijay Mukati
askIITians Faculty 2590 Points
7 years ago
Since, a²(b+c), b²(c+a), c²(a+b) ... are in A.P.
So, b²(c+a) - a²(b+c) = c²(a+b) - b²(c+a)
==> b²c + b²a - a²b - a²c = c²a + c²b - b²c - b²a
==>(b²c - a²c) + (b²a - a²b) = (c²a - b²a) + (c²b - b²c)
==>c(b² - a²) + ab(b-a) = a(c² - b²) + bc(c - b)
==>(b-a) [ c(b+a) + ab ] = (c-b) [ a(c+b) + bc ]
==> (b-a)( ab + bc ca ) = (c-b)( ab + bc + ca )

∴ Either : ab + bc + ca = 0
Or : b - a = c - b, i.e., a, b, c are in A.P.

Thanks, Vj
mycroft holmes
272 Points
7 years ago
If a^2 (b+c), b^2(c+a), c^2(a+b) are in AP
 
adding, abc to each term,
 
a^2b+a^2c+abc, b^2c+b^2a+abc, c^2a+c^2b+abc are also in AP
 
i.e. a(ab+bc+ca),b(ab+bc+ca), c(ab+bc+ca) are in AP
 
Now, if ab+bc+ca is non-zero, then dividing by ab+bc+ca, we get a,b,c are in AP

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