Grade 9Algebra
Please give answerFind the ap or gp o. ........... ....
Please give answer
Find the ap or gp o.
......
.....
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1 Answer
Arun
a(1/b+1/c) + c(1/a+1/b)
=a(c +b)/bc + c(a+b)/ab
={a²(c +b) + c²(a+b)}/abc
={a²c +a²b + c²a+c²b)}/abc
={a²c+ c²a +a²b +c²b)}/abc
={ac(a+c) +a²b +c²b)}/abc
={ac(2b) +a²b +c²b)}/abc Since a,b,c are in AP hence a+c=2b
={2abc +a²b +c²b)}/abc
=b{2ac +a² +c²)}/abc
=b{(a+c)²)}/abc
={(a+c)²)}/ac
={(a+c)(a+c))}/ac
=(2b)(a+c))}/ac Since a,b,c are in AP hence a+c=2b
=(2b)(1/c+1/a)
hence a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P if a,b,c are in AP
=a(c +b)/bc + c(a+b)/ab
={a²(c +b) + c²(a+b)}/abc
={a²c +a²b + c²a+c²b)}/abc
={a²c+ c²a +a²b +c²b)}/abc
={ac(a+c) +a²b +c²b)}/abc
={ac(2b) +a²b +c²b)}/abc Since a,b,c are in AP hence a+c=2b
={2abc +a²b +c²b)}/abc
=b{2ac +a² +c²)}/abc
=b{(a+c)²)}/abc
={(a+c)²)}/ac
={(a+c)(a+c))}/ac
=(2b)(a+c))}/ac Since a,b,c are in AP hence a+c=2b
=(2b)(1/c+1/a)
hence a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P if a,b,c are in AP
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