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
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
I believe the roots are r = 1.5 and r = 1
try those
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
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
2
To solve this problem, let's first identify the values we have:
Common difference of the arithmetic progression (a): Unknown
First term of the arithmetic progression (d): 3
Term numbers of the arithmetic progression (n1, n2, n3): 4, 6, 9
Common ratio of the geometric progression (r): Unknown
Now, we'll use the given information to form equations and solve for the unknown values.
1. For the arithmetic progression, we know that the 4th term is d + 3a, the 6th term is d + 5a, and the 9th term is d + 8a.
2. For the geometric progression, we know that the 1st term is d + 3a, the 2nd term is d + 5a, and the 3rd term is d + 8a.
We can write the equation for the geometric progression as:
(d + 5a) = (d + 3a) * r, where r is the common ratio.
Expanding this equation, we get:
d + 5a = dr + 3ar
Now, let's substitute the values of the terms from the arithmetic progression:
d + 5a = (d + 3a) * r
3 + 5a = (3 + a) * r
Expanding further, we get:
3 + 5a = 3r + ar
Rearranging the equation, we have:
5a - ar = 3r - 3
Factoring out "a" and "r" on the left side:
a(5 - r) = 3r - 3
Now, we need to find the values of "a" and "r" that satisfy this equation. We know that the first term of the arithmetic progression is 3, so we can substitute it into the equation:
a(5 - r) = 3r - 3
a(5 - r) = 3(r - 1)
Now, we have a system of equations where a(5 - r) is equal to 3(r - 1). We can compare the coefficients and solve for "a" and "r":
5 - r = 3 -- Equation 1
a = r - 1 -- Equation 2
From Equation 1, we have:
5 - r = 3
-r = 3 - 5
-r = -2
r = 2
Now, substitute the value of "r" into Equation 2:
a = r - 1
a = 2 - 1
a = 1
Therefore, the common difference of the arithmetic progression is 1, and the common ratio of the geometric progression is 2.