# if a,b,c are in H.P. then prove that a/(b+c), b/(a+c), c/(a+b) are in H.P. Please give me complete ans with steps. Thanks in advance.

SAGAR SINGH - IIT DELHI
879 Points
11 years ago

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

akshay khandelwal
18 Points
11 years ago

(b+c)/a,(a+c)/b,(a+b)/c will be in a.p.

so (b+c)/a+1,(a+c)/b+1,(a+b)/c+1 also in a.p

=(a+b+c)/a,(a+b+c)/b,(a+b+c)/c

so1/a,1/b,1/c will be in ap

so a,b,c will be in h.p

2 years ago
Dear student,

Let us assume (b+c)/a, (a+c)/b, (a+b)/c are in AP
so (b+c)/a + 1, (a+c)/b + 1, (a+b)/c + 1 also in AP
or, (a+b+c)/a, (a+b+c)/b, (a+b+c)/c are in AP
so,1/a, 1/b, 1/c are in AP
so, a, b, c are in HP

Hence, whenever a/(b + c), b/(a + c), c/(a + b) are in HP; a, b, c will also be in HP
Hence, vice-versa is also true thus when a, b, c are in HP; a/(b + c), b/(a + c), c/(a + b) will also be in HP

Thanks and regards,
Kushagra