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`        If a,b,c are in H.P then a/b+c, b/c+a, c/a+b will be inA.pH.pG.pNone`
5 months ago

Arun
22364 Points
```							Dear Anvita  Dear student,a, b , c  are in H.P   we know  that  b =  2ac /  a+ c  now  we have to prove that  a/ b+ c ,  b / c+ a , c / a + b are in h.p  i .e  we  have  to prove that  b = 2ac / a+ c ---1  $\frac{b}{c + a} = 2\left ( \frac{\frac{a}{b+ c .}\frac{c}{a+ b} }{\frac{a}{b+ c}+ \frac{c}{a+ b}} \right )$ $\frac{b}{c + a} = \frac{2ac}{a^2 + ab + bc + c^2 }$ $\frac{b}{c + a} = \frac{2ac}{a^2 + c^2 + b ( a+ c ) }$ substitute value of  a+ c  from  --1  in above equation  $\frac{b}{c + a} = \frac{2ac}{a^2 + c^2 + b (\frac{2ac}{b}) }$ $\frac{b}{c + a} = \frac{2ac}{a^2 + c^2 + 2ac }$ $\frac{b}{c + a} = \frac{2ac}{(a+ c)^{2} }$ $b = \frac{2ac}{a+ c}$ hence proved  these  are in h.p
```
5 months ago
Samyak Jain
328 Points
```							Another method : It is given that a,b,c are in H.P.By definition, 1/a , 1/b , 1/c are in A.P.Multiply each term by a+b+c(a+b+c)/a , (a+b+c)/b , (a+b+c)/c  are also in A.P.i.e. 1 + (b+c)/a , 1 + (c+a)/b , 1 + (a+b)/c  are in A.P.Subtracting 1 from each term, we get(b+c)/a , (c+a)/b , (a+b)/c  are in A.P.$\dpi{80} \therefore$ Their reciprocals a/(b+c), b/(c+a), c/(a+b) are in H.P.
```
5 months ago
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