The sum of 3numbers of a gp is 39and their product is 729. the numbers are
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the The product of the numbers is 729.
Let first term be n and common ratio be r.
Product = n*nr*nr^2=n^3.r^3=729.
n.r=(729)^(1/3)=9.
n=9/r.
n+nr+nr^2=9/r+9/r*r+9/r*r^2=39.
9/r+9+9r=39.
9/r+9r=39–9=30.
Multiplying both sides by r/3 we get 3+3r^2=10r.
3r^2–10r+3=0.
3r(r-3)-1(r-3)=0.
(3r-1)(r-3)=0.
r=1/3 or r=3.
Taking r=1/3 the numbers are 9/(1/3),9,9/(1/3)^2= 3, 9 and 27.
Taking r=3 the numbers are 9*3,9,9/3 =27,9 and 3.
Let first term be n and common ratio be r.
Product = n*nr*nr^2=n^3.r^3=729.
n.r=(729)^(1/3)=9.
n=9/r.
n+nr+nr^2=9/r+9/r*r+9/r*r^2=39.
9/r+9+9r=39.
9/r+9r=39–9=30.
Multiplying both sides by r/3 we get 3+3r^2=10r.
3r^2–10r+3=0.
3r(r-3)-1(r-3)=0.
(3r-1)(r-3)=0.
r=1/3 or r=3.
Taking r=1/3 the numbers are 9/(1/3),9,9/(1/3)^2= 3, 9 and 27.
Taking r=3 the numbers are 9*3,9,9/3 =27,9 and 3.
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