Let a_1, a_2, a_3, ldots be an A.P. If a_7 = | vedclass

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(A) Given $a_7 = a + 6d = 3$,so $a = 3 - 6d$. The product $P = a_1 a_4 = a(a + 3d) = (3 - 6d)(3 - 3d) = 18d^2 - 27d + 9$. To find the minimum,we diffe

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$24$

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(A) Given $a_7 = a + 6d = 3$,so $a = 3 - 6d$. The product $P = a_1 a_4 = a(a + 3d) = (3 - 6d)(3 - 3d) = 18d^2 - 27d + 9$. To find the minimum,we differentiate $P$ with respect to $d$: $\frac{dP}{dd} = 36d - 27 = 0 \Rightarrow d = \frac{3}{4}$. Substituting $d = \frac{3}{4}$ into $a = 3 - 6d$,we get $a = 3 - 6(\frac{3}{4}) = 3 - \frac{9}{2} = -\frac{3}{2}$. The sum of the first $n$ terms is $S_n = \frac{n}{2} [2a + (n - 1)d] = 0$. Since $n 
eq 0$,$2(-\frac{3}{2}) + (n - 1)(\frac{3}{4}) = 0 \Rightarrow -3 + \frac{3(n - 1)}{4} = 0 \Rightarrow \frac{3(n - 1)}{4} = 3 \Rightarrow n - 1 = 4 \Rightarrow n = 5$. We need to calculate $n! - 4 a_{n(n+2)} = 5! - 4 a_{5(7)} = 120 - 4 a_{35}$. $a_{35} = a + 34d = -\frac{3}{2} + 34(\frac{3}{4}) = -\frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$. Thus,$120 - 4(24) = 120 - 96 = 24$.

Let $a_1, a_2, a_3, \ldots$ 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 :

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