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