Asked by Tara
The first, second and third terms of a geometric progression are 2k+3, k+6 and k, respectively. Given that all the terms of geometric progression are positive, calculate
(a) the value of the constant k
(b) the sum to infinity of the progression.
(a) the value of the constant k
(b) the sum to infinity of the progression.
Answers
Answered by
Steve
since the ratio between terms is constant,
(k+6)/(2k+3) = k/(k+6)
(k+6)^2 = k(2k+3)
k^2+12k+36 = 2k^2+3k
k^2 - 9k - 36 = 0
(k-12)(k+3) = 0
k = 12
and the sequence is
27,18,12,...
so the ratio is 2/3 and S = 27/(1/3) = 81
(k+6)/(2k+3) = k/(k+6)
(k+6)^2 = k(2k+3)
k^2+12k+36 = 2k^2+3k
k^2 - 9k - 36 = 0
(k-12)(k+3) = 0
k = 12
and the sequence is
27,18,12,...
so the ratio is 2/3 and S = 27/(1/3) = 81
Answered by
PAUL
IF THE 6TH TERM OF A SERIES IS 24 AND THE COMMON DIFFERENCE IS 3 FIND THE SERIES
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