In an A.P., the sixth term a 6 = 2 . If the product a 1 a 4 a 5 is the greatest, then the common difference of the A.P. is equal to [2024]

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Solution

Q.
In an A.P., the sixth term $a_{6} = 2 .$ If the product $a_{1} a_{4} a_{5}$ is the greatest, then the common difference of the A.P. is equal to [2024]

1 $\frac{8}{5}$

2 $\frac{5}{8}$

3 $\frac{3}{2}$

4 $\frac{2}{3}$

Ans.
(1)

We have, $a_{6} = 2 \Rightarrow a + 5 d = 2 \Rightarrow d = \frac{2 - a}{5}$

Let $a_{1} a_{4} a_{5} = λ$

$\Rightarrow λ = a (a + 3 d) (a + 4 d)$ $= a (a^{2} + 7 a d + 12 d^{2})$ $= a^{3} + 7 a^{2} d + 12 a d^{2}$

$= a^{3} + 7 a^{2} (\frac{2 - a}{5}) + 12 a {(\frac{2 - a}{5})}^{2}$ $[∵ d = \frac{2 - a}{5}]$

$= a^{3} + \frac{14}{5} a^{2} - \frac{7}{5} a^{3} + \frac{12 a}{25} (4 - 4 a + a^{2})$

$= a^{3} + \frac{14}{5} a^{2} - \frac{7}{5} a^{3} + \frac{48}{25} a - \frac{48}{25} a^{2} + \frac{12}{25} a^{3}$

$= \frac{2}{25} a^{3} + \frac{22}{25} a^{2} + \frac{48}{25} a = \frac{2}{25} (a^{3} + 11 a^{2} + 24 a)$

$∴$ $\frac{d λ}{d a} = \frac{2}{25} (3 a^{2} + 22 a + 24)$

$λ$ to be greatest

$∴$ $\frac{d λ}{d a} = 0 \Rightarrow 3 a^{2} + 22 a + 24 = 0 \Rightarrow a = - 6, - \frac{4}{3}$

$\frac{d^{2} λ}{d a^{2}} = \frac{2}{25} (6 a + 22), For a = - 6, \frac{d^{2} λ}{d a^{2}} < 0$

So, $λ$ is maximum

For $a = \frac{- 4}{3}, \frac{d^{2} λ}{d a^{2}} > 0, λ$ is minimum $∴$ $d = \frac{2 - (- 6)}{5} = \frac{8}{5}$

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