JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 7)
Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the
product a1a4a5, is :
$${3 \over 2}$$
$${6 \over 5}$$
$${8 \over 5}$$
$${2 \over 3}$$
Explanation
first term = a, Common difference = d
$$ \therefore $$ a + 5d = 2
a1. a4. a5 = a(a + 3d) (a + 4d)
f(d) = (2 – 5d) (2 – 2d) (2 – d)
$$ \Rightarrow $$ $$f'(d) = 0 \Rightarrow d = {2 \over 3},{8 \over 5}$$
$$ \Rightarrow $$ $$f''(d) < 0\,at\,d = {8 \over 5}$$
$$\, \Rightarrow d = {8 \over 5}$$
$$ \therefore $$ a + 5d = 2
a1. a4. a5 = a(a + 3d) (a + 4d)
f(d) = (2 – 5d) (2 – 2d) (2 – d)
$$ \Rightarrow $$ $$f'(d) = 0 \Rightarrow d = {2 \over 3},{8 \over 5}$$
$$ \Rightarrow $$ $$f''(d) < 0\,at\,d = {8 \over 5}$$
$$\, \Rightarrow d = {8 \over 5}$$
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