plss give answer of question 6 and 7
lmp -- question 6
plsssssss solve
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hi friend here is your required answer and solution of 7 is give below
we have AM greater than=GM and GM=b^2=ac
so b^3=abc=4(from the question)
b=4^(1/3)=2^(2/3)
so the minimum values b can take such that a,b,c are in AP is 2^(2/3)
Ans isB
we have AM greater than=GM and GM=b^2=ac
so b^3=abc=4(from the question)
b=4^(1/3)=2^(2/3)
so the minimum values b can take such that a,b,c are in AP is 2^(2/3)
Ans isB
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Feliciano9:
tq for solving
thanks for selectind as brainlist answer
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