The sixth term of an A.P. is 2, its common difference is greater than one. Find the value of the common difference so that the product of the first, fourth and fifth terms is the greatest.


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Question Number 108854 by ZiYangLee last updated on 19/Aug/20

$The sixth term of an A . P . is 2, its$ $common difference is greater than$ $one . Find the value of the common$ $difference so that the product of the$ $first, fourth and fifth terms is the$ $greatest .$
	Answered by $@y@m last updated on 20/Aug/20
	$Let a be the first term and x be the common difference. a+5x=2 a=2−5x.....(1) Let y=a(a+3x)(a+4x) ⇒y=(2−5x)(2−2x)(2−x) {from (1) ..........(√)(2) =(4−14x+10x^2 )(2−x) =8−4x−28x+14x^2 +20x^2 −10x^3 y=−10x^3 +34x^2 −32x+8 .....(3) (dy/dx)=−30x^2 +68x−32 ....(4) For maxima or minima, (dy/dx)=0 ⇒−30x^2 +68x−32=0 ⇒15x^2 −34x+16=0 x=((34±(√(1156−960)))/(30)) x=((34±14)/(30))= ((48)/(30)), ((20)/(30)) =(8/5), (2/3) ATQ, x>1 ∴x=(8/5) Now, (d^2 y/dx^2 ) = −60x+68 Now, (d^2 y/dx^2 ) _(at x=(8/5)) =−60×(8/5)+68=−28<0 ∴ y is maximum when x=(8/5)=1.6$
	$L e t a b e t h e f i r s t t e r m a n d$ $x b e t h e c o m m o n d i f f e r e n c e .$ $a + 5 x = 2$ $a = 2 - 5 x . . . . . (1)$ $L e t y = a (a + 3 x) (a + 4 x)$ $\Rightarrow y = (2 - 5 x) (2 - 2 x) (2 - x) {f r o m (1)$ $. . . . . . . . . . \sqrt{} (2)$ $= (4 - 14 x + 10 x^{2}) (2 - x)$ $= 8 - 4 x - 28 x + 14 x^{2} + 20 x^{2} - 10 x^{3}$ $y = - 10 x^{3} + 34 x^{2} - 32 x + 8 . . . . . (3)$ $\frac{d y}{d x} = - 30 x^{2} + 68 x - 32 . . . . (4)$ $F o r m a x i m a o r m i n i m a,$ $\frac{d y}{d x} = 0$ $\Rightarrow - 30 x^{2} + 68 x - 32 = 0$ $\Rightarrow 15 x^{2} - 34 x + 16 = 0$ $x = \frac{34 \pm \sqrt{1156 - 960}}{30}$ $x = \frac{34 \pm 14}{30} = \frac{48}{30}, \frac{20}{30} = \frac{8}{5}, \frac{2}{3}$ $A T Q, x > 1$ $∴ x = \frac{8}{5}$ $N o w, \frac{d^{2} y}{{d x}^{2}} = - 60 x + 68$ $N o w, \frac{d^{2} y}{{d x}^{2}}_{a t x = \frac{8}{5}} = - 60 \times \frac{8}{5} + 68 = - 28 < 0$ $∴ y i s m a x i m u m w h e n x = \frac{8}{5} = 1.6$
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