Asked by sherif
the first termn of an exponential sequence (GP) is greater than the fourth term by 13.5 and the fourth is greater than the third by 9.Find:the common ratio and the first term.
Answers
Answered by
Steve
a = ar^3 + 13.5
ar^3 = ar^2 + 9
a(1-r^3) = 13.5
ar^2(r-1) = 9
Now divide both sides to get
(1-r^3)/(r^3-r^2) = 13.5/9
2-2r^3 = 3r^3-3r^2
5r^3-3r^2-2 = 0
Hmmm. No real solutions.
ar^3 = ar^2 + 9
a(1-r^3) = 13.5
ar^2(r-1) = 9
Now divide both sides to get
(1-r^3)/(r^3-r^2) = 13.5/9
2-2r^3 = 3r^3-3r^2
5r^3-3r^2-2 = 0
Hmmm. No real solutions.
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