ONLY FOR MATHS ARYABHATTA,GENIUS AND UPPER RANKERS
If a,b,c are in A.P . Prove that the following is in A.P with the most suitable method

That dot(.) is comma(,)
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Answers
Answered by
3
Hey there,
There are various methods for solving this.
METHOD--1
IF a,b,c are in AP
=>a/abc , b/abc ,c/abc are also in AP
=>1/bc ,1/ac. ,1/ab. Are in AP
=>(ab+bc+ca)/bc ,(ab+bc+ca)/ac ,
(ab+bc+ca)/ab are also in AP
=>1+(ab+ca)/bc ,1+(ab+bc)/ac,
1+(bc+ca)/ab are also in AP.
=>(ab+ca)/bc ,(ab+bc)/ac ,(bc+ca)/ab
are also in AP.
=>a(b+c)/bc ,b(a+c)/ac, c(a+b)/ab are
also in AP.
=>a^2(b+c),b^2(a+c),c^2(a+b) are also
in AP[multiplying by abc]
Hope it helps.
There are various methods for solving this.
METHOD--1
IF a,b,c are in AP
=>a/abc , b/abc ,c/abc are also in AP
=>1/bc ,1/ac. ,1/ab. Are in AP
=>(ab+bc+ca)/bc ,(ab+bc+ca)/ac ,
(ab+bc+ca)/ab are also in AP
=>1+(ab+ca)/bc ,1+(ab+bc)/ac,
1+(bc+ca)/ab are also in AP.
=>(ab+ca)/bc ,(ab+bc)/ac ,(bc+ca)/ab
are also in AP.
=>a(b+c)/bc ,b(a+c)/ac, c(a+b)/ab are
also in AP.
=>a^2(b+c),b^2(a+c),c^2(a+b) are also
in AP[multiplying by abc]
Hope it helps.
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