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 :
JEE · Math · previous-year question
- A.$${3 \over 2}$$
- B.$${6 \over 5}$$
- C.$${8 \over 5}$$correct
- D.$${2 \over 3}$$
Answer
C. $${8 \over 5}$$
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) $$ \Rightarrow $$ $$f'(d) = 0 \Rightarrow d = {2 \over 3},{8 \over 5}$$ $$ \Rightarrow $$ $$f''(d) < 0\,at\,d = {8 \over 5}$$ $$\, \Rightarrow d = {8 \over 5}$$
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