Verified Solution • 2025–26 Session moderate

QJEEM23Mae36d: JEE Main 2023 Mathematics - Class 11 JEEM23Mae36d Solution: Let be an A.P. If , the product is minimum and the sum of its…

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be an A.P. If a7=3a_7=3 , the product a1a4a_1 a_4 is minimum and the sum of its first nn terms is zero, then n!−4an(n+2)n !-4 a_{n(n+2)} is equal to :

Multiple Choice Options

  • A 24
  • B 3814\frac{381}{4}
  • C 9
  • D 334\frac{33}{4}

Step-by-Step Verified Solution

a7=3⇒a+6d=3⇒a=3−6da_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d

amp;a1⋅a4=a(a+3d)amp;⇒(3−6d)(3−6d+3d)amp;⇒3(1−2d)3(1−d)amp;⇒9(2d2−3d+1)\begin{aligned}& a_{1} \cdot a_{4}=a(a+3 d) \\\\& \Rightarrow(3-6 d)(3-6 d+3 d) \\\\& \Rightarrow 3(1-2 d) 3(1-d) \\\\& \Rightarrow 9\left(2 d^{2}-3 d+1\right)\end{aligned}

Let f(d)=2d2−3d+1f(d)=2 d^{2}-3 d+1

f′(d)=4d−3⇒d=34f^{\prime}(d)=4 d-3 \Rightarrow d=\frac{3}{4}

∴a=3−6⋅34=3−92=−32\therefore a=3-6 \cdot \frac{3}{4}=3-\frac{9}{2}=-\frac{3}{2}

Sn=0S_{n}=0

n2(29+(n−1)d)=0\frac{n}{2}(29+(n-1) d)=0

⇒2⋅(−32)+(n−1)(34)=0\Rightarrow 2 \cdot\left(-\frac{3}{2}\right)+(n-1)\left(\frac{3}{4}\right)=0

⇒3=34(n−1)\Rightarrow \quad 3=\frac{3}{4}(n-1)

⇒n=5\Rightarrow n=5

Now, n!−4⋅an(n+2)n !-4 \cdot a_{n(n+2)}

amp;=5!−4⋅a35amp;=120−4(−32+34⋅34)amp;=120−(−6+102)amp;=120−(96)amp;=24\begin{aligned}& =5 !-4 \cdot a_{35} \\\\& =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\\\& =120-(-6+102) \\\\& =120-(96) \\\\& =24\end{aligned}

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