JEE MAIN - Mathematics (2017 - 9th April Morning Slot - No. 1)
If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :
2
4$${^{{1 \over 3}}}$$
4$${^{{2 \over 3}}}$$
4
Explanation
a, b and c are in AP.
$$ \therefore $$ a + c = 2b
As, abc = 8
$$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8
$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4
$$ \therefore $$ ac = 4 and a + c = 4
Then,
b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2
$$ \therefore $$ a + c = 2b
As, abc = 8
$$ \Rightarrow $$ac$$\left( {{{a + c} \over 2}} \right)$$= 8
$$ \Rightarrow $$ ac(a + c) = 16 = 4 $$ \times $$ 4
$$ \therefore $$ ac = 4 and a + c = 4
Then,
b = $$\left( {{{a + c} \over 2}} \right)$$ = $${4 \over 2}$$ = 2
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