If a,b,c are in A.P the show that a (b+c)/bc , b (c+a)/ca and c (a+b)/ab are also in the A.P
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a is in ap.
a(b+c)/bc =0
a*b + a*c/b*c =0
2a/1=0
2a=0
a=0*2
a=0
therefore a is in ap
given b is in ap
b(c+a)/ca=0
bc+ba/c*a=0
2b=0
b=0
b is in ap
given c is in ap
c(a+b)/ab=0
ca+cb/a*b=0
2c=0
c=0.
therefore c is in ap
a(b+c)/bc =0
a*b + a*c/b*c =0
2a/1=0
2a=0
a=0*2
a=0
therefore a is in ap
given b is in ap
b(c+a)/ca=0
bc+ba/c*a=0
2b=0
b=0
b is in ap
given c is in ap
c(a+b)/ab=0
ca+cb/a*b=0
2c=0
c=0.
therefore c is in ap
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