Find three numbers in G.P. whose sum is 28 and whose product is 512.

1.	Find three numbers in G.P. whose sum is 28 and whose product is 512.
Answer» Let the three numbers be\xa0a,ar,\xa0ar2,where\xa0r\xa0is the common ratio.⇒a+ar+ar2=28\xa0and\xa0a3r3=512⇒ar=8⇒a+ar2=20⇒8r2−20r+8=0⇒r=2,r=21\u200bIf\xa0r=2,a=4Therefore, the three numbers are\xa04,8,16<br>The three numbers in GP are 4 , 8 , 16.\tLet the three numbers in GP be a/r , a , ar.\tNow the product of three numbers is given as 512.a/r × a × ar = 512a³ = 512a = 8\tNow sum of the three numbers is given as 28.a/r + a + ar = 288/r + 8 + 8r = 288 + 8r² = 20r8r² - 20r + 8 =08r² - 16r - 4r + 8 =08r(r-2) - 4(r-2) = 0(r-2)(8r-4) = 0\tTherefore, r = 2 or r=1/2\tNow , the three numbers are 4,8,16.

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