JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(a_1, a_2, a_3, \ldots \ldots\) 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 :
- A \(24\)
- B \(\frac{33}{4}\)
- C \(\frac{381}{4}\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(24\)
Step-by-step Solution
Detailed explanation
\(a +6 d =3\) \(Z = a ( a +3 d )\) \(=(3-6 d )(3-3 d )\) \(=18 d ^2-27 d +9\) Differentiating with respect to d \(\Rightarrow 36 d -27=0\) \(\Rightarrow d =\frac{3}{4}, \text { from (1) } a =\frac{-3}{2},( Z =\text { minimum) }\)…
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