Let a _ 1 , a _ 2 , a _ 3 , . be an A.P. If a _ 7 = 3 , the product...

We know the ${n}^{th}$ term of an A.P. is given by,

${a}_{n} = a + \left(n - 1\right)d$

Given, ${a}_{7} = 3$

$⇒ a + 6d = 3$

$⇒ a = 3 - 6d$

And, ${a}_{1}{a}_{4} = a\left(a + 3d\right)$

$= \left(3 - 6d\right) \left(3 - 3d\right)</p><p>$= 18{d}^{2} - 27d + 9$</p><p>Given product \left({a}_{1}{a}_{4}\right) is minimum then,</p><p>Let$f ( d ) = 18{d}^{2} - 27d + 9$</p><p>${f}^{'} ( d ) = 36d - 27$</p><p>Product to be minimum,${f}^{'}\left(d\right)$= 0$

$⇒ 36d - 27 = 0$

$⇒ d =$\dfrac{27}{36}=\dfrac{3}{4}

So, $a = 3 -$\dfrac{9}{2} = \dfrac{- 3}{2}

Given, ${S}_{n} = 0$

${S}_{n}$= \dfrac{n}{2}\left[2a +\left(n - 1\right)d\right] = 0

$- 3 + \left(n - 1\right)\dfrac{3}{4} =$ 0$</p><p>$⇒ n = 5$</p><p>Now$n ! - 4{a}_{n ( n + 2 )} = 5 ! - 4{a}_{35}$</p><p>$= 120 - 4 \left(a + 34d\right)

$= 120 - 4 \left(\dfrac{- 3}{2} +34 \times \dfrac{3}{4}\right)$

$= 120 + 6 - 102 = 24$