JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Application of Derivatives

Let

a_{1}, a_{2}, a_{3}, \dots\dots

be and A.P with

a_{6} = 2

. Then the common difference of this A.P., which maximizes the product

a_{1} a_{4} a_{5}

Answer: (B)

Solution

Write the terms through the sixth,

a_{6} = a + 5 d = 2

, so

a = 2 - 5 d

. Then

a_{1} a_{4} a_{5} = (2 - 5 d) (2 - 2 d) (2 - d)

, a cubic in

d

. Differentiating and setting the derivative to zero gives

15 d^{2} - 34 d + 16 = 0

, whose roots are

d = \frac{2}{3}

and

d = \frac{8}{5}

. The second derivative

68 - 60 d

is negative at

d = \frac{8}{5}

and positive at

d = \frac{2}{3}

, so the product is maximised when the common difference is

\frac{8}{5}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option