find three nos in AP whose product is equal to the square of their sum.


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Question Number 31281 by MURALI last updated on 05/Mar/18

$f i n d t h r e e n o s i n A P w h o s e p r o d u c t$ $i s e q u a l t o t h e s q u a r e o f t h e i r s u m .$
	Answered by Rasheed.Sindhi last updated on 07/Mar/18
	$Let the middle number is a and the common difference is d ∴ Numbers are a−d,a,a+d (a−d)(a)(a+d)={(a−d)+(a)+(a+d)}^2 a(a^2 −d^2 )=(3a)^2 a^3 −9a^2 −ad^2 =0 a(a^2 −9a−d^2 )=0 ^(•1) a=0 ∣ ^(•2) a^2 −9a−d^2 =0 ^(•1) As a=0 is free of d so d may be any number. Hence in general −d,0,d are required nos. For example −3,0,3 ; −4,0,4 etc. ^(•2) a^2 −9a−d^2 =0 a=((9±(√(81+4d^2 )))/2) In general ((9±(√(81+4d^2 )))/2)−d , a , ((9±(√(81+4d^2 )))/2)+d are required numbers for any number d. If a,d∈Z , d=±6, ±20 work. For d=6 required numbers are: 6,12,18 or −9,−3 , 3 For d=20 required numbers are: 5,25, 45 or −36,−16, 4$
	$Let the middle number is a$ $and the common difference is d$ $∴ Numbers are a - d, a, a + d$ $(a - d) (a) (a + d) = {(a - d) + (a) + (a + d)}^{2}$ $a (a^{2} - d^{2}) = {(3 a)}^{2}$ $a^{3} - {9 a}^{2} - {ad}^{2} = 0$ $a (a^{2} - 9 a - d^{2}) = 0$ $^{∙ 1} a = 0 ∣^{∙ 2} a^{2} - 9 a - d^{2} = 0$ $^{∙ 1} As a = 0 is free of d so d may be any number .$ $Hence in general - d, 0, d are required nos .$ $For example - 3, 0, 3; - 4, 0, 4 etc .$ $^{∙ 2} a^{2} - 9 a - d^{2} = 0$ $a = \frac{9 \pm \sqrt{81 + {4 d}^{2}}}{2}$ $In general$ $\frac{9 \pm \sqrt{81 + {4 d}^{2}}}{2} - d, a, \frac{9 \pm \sqrt{81 + {4 d}^{2}}}{2} + d$ $are required numbers for any number d .$ $If a, d \in Z, d = \pm 6, \pm 20 work .$ $For d = 6 required numbers are :$ $6, 12, 18 or - 9, - 3, 3$ $For d = 20 required numbers are :$ $5, 25, 45 or - 36, - 16, 4$
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