Question

The 4th, 6th and 9th term of an arithmetical progression form the first three times of a geometric progression. If the first term of the arithmetic progression is 3 determine the common difference of the arithmetic progression and common ratio of the geometric al progression

Answer 1

4 th ==> 3+3d = a
6 th ==> 3 + 5d = a r
9 th ==> 3 + 8 d = a r^2
--------------------------------------
15 + 15 d = 5 a
9 + 15 d = 3 a r
-------------------------- subtract
6 = a (5-3r)
---------------------------
24 + 40 d = 8 a r
15 + 40 d = 5 a r^2
----------------------------- subtract
9 = a ( 8 r - 5 r^2)
so
9 / ( 8 r - 5 r^2) = 6 / (5-3r)
9 (5 - 3 r) = 6 (8 r - 5 r^2)
45 - 27 r = 48 r - 30 r^2
30 r^2 - 75 r + 45 = 0

Answer 2

I believe the roots are r = 1.5 and r = 1
try those

Answer 3

if r = 1
6 = a (5-3r)
6 = a (2)
a = 3
a, a, a ------- boring, forget it
if r = 1.5
6 = a (5-3r) = 5 a - 4.5 a = .5 a
a = 12
12 , 18 , 27 first three geometric
3+3d = a
3 d = 12 - 3 = 9
d = 3
3,6,9,12 YES !!! the fourth arith is the first geo

Answer 4

from the given:
(a+5d)/(a+3d) = (a+8s)/(a+5d)
but a = 3
(3+5d)^2 = (3+3d)(3+8d)
9 + 30d + 25d^2 = 9 + 33d + 24d^2
d^2 - 3d = 0
d(d-3) = 0
d = 0, or d = 3

so for the AP, when d = 3
term(4) = 3 + 3(3) =12
term(6) = 3 + 5(3) = 18
term(9) = 3 + 8(3) = 27

sure enough: 18/12 = 3/2, and 27/18 = 3/2, so they are also in a GP
so r = 3/2

if d = 0,
term(4), term(6), and term(9) are all equal to 3
so 3,3,3 technichally would be a GP with r = 1

Answer 5

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