Question
in a geometric progression, the first term is a and the common ratio is r. the sum of the first two terms is 12 and the third term is 16.
a) determine the ratio (ar^2)/a+ar( aready found the answer as 4/3)
b) if the first term is larger than the second term, find the value of r.
a) determine the ratio (ar^2)/a+ar( aready found the answer as 4/3)
b) if the first term is larger than the second term, find the value of r.
Answers
mathhelper
a + ar = 12
ar^2 = 16
ar^2/(a(1 + r)) = 16/12
r^2/(1+r) = 4/3 <==== you had that
3r^2 = 4 + 4r
3r^2 - 4r - 4 = 0
(r-2)(3r+2) = 0
r = 2 or r = -2/3
if r = 2,
ar^2 = 16 , a = 4 , and the sequence is 4, 8, 16, ...
if r = -2/3,
a(4/9) = 16, a = 36, and the sequence is 36, -24, 16,
but it said, term1 > term2, so it must be the 2nd sequence, where
r = -2/3
ar^2 = 16
ar^2/(a(1 + r)) = 16/12
r^2/(1+r) = 4/3 <==== you had that
3r^2 = 4 + 4r
3r^2 - 4r - 4 = 0
(r-2)(3r+2) = 0
r = 2 or r = -2/3
if r = 2,
ar^2 = 16 , a = 4 , and the sequence is 4, 8, 16, ...
if r = -2/3,
a(4/9) = 16, a = 36, and the sequence is 36, -24, 16,
but it said, term1 > term2, so it must be the 2nd sequence, where
r = -2/3
Orji esther
No idea
Juma
Sum of 2nd term and 3rd term is 12 and sum of 4th term and 5th term is 300 in g.p .find first term and common ratio