9put a=1 ,b=2 ,c=3
u find that they form an ap!
later on u go to prove that they are always in ap by checkin if
bc-a2+ab-c2 =2(ac -b2)
if a,b,c are in ap then bc-a2,ca-b2,ab-c2 are in
a.ap
b.gp
c.hp
d.none
plz show ur solution
also give some general guidelines for solving such probs..............as i face trouble sometimes with them!
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5 Answers
9
341I have just got hammered in another site for a subjective answer, but I am still gonna stick my neck out;
a,b,c are in A.P. implies
a(a+b+c) , b(a+b+c), c(a+b+c) are in A.P.
or
a2+ab+ac, b2+ab+bc, c2+ac + bc are in A.P.
which implies that
a2+ab+ac -(ab+bc+ac), b2+ab+bc -(ab+bc+ac), c2+ac + bc (ab+bc+ac) are in A.P.
or
a2-bc, b2-ac, c2-ab are in A.P.
1@prophet sir.
thank you
can u tell the basic thrust behind solving these problems?
341If a,b,c are in A.P., ka, kb, kc are also in A.P. and so are a+k, b+k, c+k.
Here, first I subtracted the first two terms and got (a-b)(a+b+c), so I took multiplying a,b,c by a+b+c as a starting point. Then it was obvious how to get to the final configuration
9dude the basic thrust behind solving these problems is pattern recognition
u need to solve a lot of probs to gain experience