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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

```
4 years ago Vijay Mukati
2590 Points
```							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 = 0Or : b - a = c - b, i.e., a, b, c are in A.P.Thanks, Vj
```
4 years ago
```							If  are in AP adding, abc to each term,  are also in AP i.e.  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
```
4 years ago
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