find the three number of GP whose sum is 19 and product is 216
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Answered by
138
Let three numbers be a/r ,a and ar .
a/r+a+ar=19......(1)
a/r×a×ar=216
a^3=216
a=6.......(2)
from 1). 6/r+6+6r=19
6r^2-13r+6=0
6r^2-9r-4r+6=0
3r(2r-3)-2(2r-3)=0
(3r-2)(2r-3)=0
so r=2/3,3/2.
case1) when r=2/3,a=6 then numbers :9,6,4
case2) : when r=3/2,a=6 then numbers :4,6,9
a/r+a+ar=19......(1)
a/r×a×ar=216
a^3=216
a=6.......(2)
from 1). 6/r+6+6r=19
6r^2-13r+6=0
6r^2-9r-4r+6=0
3r(2r-3)-2(2r-3)=0
(3r-2)(2r-3)=0
so r=2/3,3/2.
case1) when r=2/3,a=6 then numbers :9,6,4
case2) : when r=3/2,a=6 then numbers :4,6,9
Answered by
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◆ Geometric Progressions ◆
Hey !!
Check the attachment.
Hope it helps you :)
Hey !!
Check the attachment.
Hope it helps you :)
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