If a, b,c are in A. P show that a^2(b+c), b^2(c+a),c^2(a+b) are also in A. P
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1) if a2(b+c) , b2(a+c), c2(a+b) are in AP then
2b2(a+c) = a2(b+c) + c2(a+b) (we have to prove this) ..............................1
RHS:
= a2 (b+c) + c2 (a+b)
= a2b +a2c +c2a +c2b
=b(a2 +c2) + ac(a+c)
since a,b,c are in AP so b=a+c/2
RHS = (a+c)(a2 +b2)/2 + ac(a+c)
=(a+c)(a2 +b2 +2ac)/2
=(a+c)3 /2 or 2(a+c) b2 (after putting a+c =b)
hence proved
2) (ab+ac)/bc , (bc+ba)/ac ,(ca+bc)/ab are in AP so
2(bc+ba)/ac = (ab+ac)/bc + (ac+bc)/ab we have to prove this
multiplying the equation by abc
now we get
2b2(a+c) = a2(b+c) + c2(a+b) we have to prove this
this expression is same as of eq 1 of previous ans ,so now take RHS and prove as i have done in previous ans...
Please mark it as brainiest answer .........
2b2(a+c) = a2(b+c) + c2(a+b) (we have to prove this) ..............................1
RHS:
= a2 (b+c) + c2 (a+b)
= a2b +a2c +c2a +c2b
=b(a2 +c2) + ac(a+c)
since a,b,c are in AP so b=a+c/2
RHS = (a+c)(a2 +b2)/2 + ac(a+c)
=(a+c)(a2 +b2 +2ac)/2
=(a+c)3 /2 or 2(a+c) b2 (after putting a+c =b)
hence proved
2) (ab+ac)/bc , (bc+ba)/ac ,(ca+bc)/ab are in AP so
2(bc+ba)/ac = (ab+ac)/bc + (ac+bc)/ab we have to prove this
multiplying the equation by abc
now we get
2b2(a+c) = a2(b+c) + c2(a+b) we have to prove this
this expression is same as of eq 1 of previous ans ,so now take RHS and prove as i have done in previous ans...
Please mark it as brainiest answer .........
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