Question

The sum of three numbers of an A P is 18 and their product is 192. Find the numbers

Answer 1

hope it's helpful to u...☺☺

Answer 2

Answer:

$Required\:three \: number\\are \:(4,6,8)\:Or\:(8,6,4)$

Step-by-step explanation:

$Let \: (a-d),a,(a+d)\:are \\three \: consecutive\: terms\\of \:A.P$

According to the problem given,

$sum \:of \: numbers=18$

$\implies a-d+a+a+d=18$

$\implies 3a=18$

$\implies a = \frac{18}{3}=6$

$Product\:of \: numbers=192$

$\implies (a-d)a(a+d)=192$

$\implies (a^{2}-d^{2}a=192$

$\implies (6^{2}-d^{2})\times 6=192$

$\implies 36-d^{2}=\frac{192}{6}$

$\implies 36-d^{2}=32$

$\implies -d^{2}=32-36$

$\implies -d^{2}=-4$

$\implies d^{2}=4$

$\implies d = ±\sqrt{2^{2}}$

$\implies d = ± 2$

$Case\: 1 \\If \:a = 6,\:d=2$

$The\: numbers\:are ,\\(a-d),a,(a+d)\\(6-2),6,(6+2)\\4,6,8$

$Case \:2 \\If \:a=6,\:d=-2\\Three \: numbers \:are ,\\8,6,4$

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