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```        a,b,c are ub A.P. the prove that

(i)a^2(b+c), b^2(c+a), c^2(a+b) are in A.P.
(ii) (ab+ac)/bc, (bc+ba)/ca, (ca+bc)/ab are in A.P.```
6 years ago

510 Points
```										1)  if a2(b+c) , b2(a+c), c2(a+b)  are in AP then
2b2(a+c)  =  a2(b+c) + c2(a+b)        (we have to prove this)        ..............................1
RHS:
= a2 (b+c) + c2 (a+b)
= a2b +a2c +c2a +c2b
=b(a2 +c2)   +  ac(a+c)
since a,b,c are in AP so b=a+c/2
RHS =     (a+c)(a2 +b2)/2 + ac(a+c)
=(a+c)(a2 +b2 +2ac)/2
=(a+c)3 /2           or   2(a+c) b2                 (after putting a+c =b)
hence proved

2)      (ab+ac)/bc , (bc+ba)/ac ,(ca+bc)/ab are in AP so
2(bc+ba)/ac    =    (ab+ac)/bc + (ac+bc)/ab          we have to prove this
multiplying the equation by abc
now we get
2b2(a+c) = a2(b+c) + c2(a+b)               we have to prove this
this expression is same as of eq 1 of previous ans ,so now  take RHS and prove as i have done in previous ans...

```
6 years ago
mycroft holmes
266 Points
```										Two results to remember:
If k is a constant and a,b,c are in AP, the following are also in AP-
1) ka,kb,kc

2) a+k, b+k, c+k

Using this,

```
6 years ago
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