Let a_1,a_2,a_3, ... be an A.P. If a_7 = 3, the product a_1a_4 is minimum and the sum of its first n terms is zero, then n! - 4a_n(n + 2) is equal to

(d) : Given, a_7=3 => a+6 d=3,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+9Differentiating with respect to d,\ => 36 d-27=0 => d=(3 / 4),( From (1), a=(-3 / 2),(Z= minimum )Now, S_n=(n / 2)[2 a+(n-1) d]=(n / 2)(-3+(n-1) (3 / 4))=0n(3 n-15)=0 => n=5Now, n!-4 a_n(n+2)=5!-4 a_35=120-4(a_35)\ =120-4(a+(35-1) d) \ =120-4((-3 / 2)+34((3 / 4))) =120-4((-6+102 / 4))=120-96=24\

Let a_{1},a_{2},a_{3}, be an A.P. If a_{7} = 3, the ... | Sequence & Series (Mathematics)

(d) : Given,

a_7=3

\Rightarrow a+6 d=3,

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

\begin{aligned}& \Rightarrow 36 d-27=0 \Rightarrow d=\frac{3}{4},(\text { From }(1), \\& a=\frac{-3}{2},(Z=\text { minimum })\end{aligned}

Now,

S_n=\frac{n}{2}[2 a+(n-1) d]=\frac{n}{2}\left(-3+(n-1) \frac{3}{4}\right)=0

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

Now,

n!-4 a_{n(n+2)}=5!-4 a_{35}=120-4\left(a_{35}\right)

\begin{aligned}& =120-4(a+(35-1) d) \\ &=120-4\left(\frac{-3}{2}+34\left(\frac{3}{4}\right)\right) \\& =120-4\left(\frac{-6+102}{4}\right)=120-96=24\end{aligned}

Explanation

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