Let a 1 , a 2 , a 3 , . . . . be an A.P. If a 7 = 3 , the product a 1 a 4 is minimum and the sum of its first n terms is zero, then n - 4 a n ( n + 2 ) is equal to [2023]

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Solution

Q.
Let $a_{1}, a_{2}, a_{3}, . . . .$ be an A.P. If $a_{7} = 3,$ the product $a_{1} a_{4}$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4 a_{n (n + 2)}$ is equal to [2023]

1 381/4

2 9

3 33/4

4 24

Ans.
(4)

$Given, a_{7} = 3$

$\Rightarrow a + 6 d = 3 \dots (1)$

$Let Z = a_{1} a_{4} = a (a + 3 d) = (a + 6 d - 6 d) (a + 6 d - 3 d)$

$= (3 - 6 d) (3 - 3 d) = 18 d^{2} - 27 d + 9$

$Differentiating with respect to d,$

$\Rightarrow 36 d - 27 = 0 \Rightarrow d = \frac{3}{4}$ , $From (1), a = \frac{- 3}{2} (Z is minimum)$

$Now, S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [- 3 + (n - 1) \frac{3}{4}] = 0$

$n (3 n - 15) = 0 \Rightarrow n = 5$

$Now, n! - 4 a_{n (n + 2)} = 5! - 4 a_{35} = 120 - 4 (a_{35})$

$= 120 - 4 (a + (35 - 1) d) = 120 - 4 (\frac{- 3}{2} + 34 (\frac{3}{4}))$

$= 120 - 4 (\frac{- 6 + 102}{4}) = 120 - 96 = 24$

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