The first two terms of a geometric sequence and an arithmetic sequence are the same. The first term is 12. The sum of the first three terms of the geometric sequence is 3 more than the sum of the first three terms of the arithmetic sequence. Determine TWO possible values for the common ratio, r, of the geometric sequence
Geometric sequence = a,ar,arē,__,__,...
Arithmetic sequence= a,a+d,a+2d,__,__,...
The first two terms of a geometric sequence and an arithmetic sequence are the same.a = a, ar = a+d
The first term is 12.
a = 12, 12r = 12+d
The sum of the first three terms of the geometric sequence is 3 more than the sum of the first three terms of the arithmetic sequence. (a)+(ar)+(arē) = (a)+(a+d)+(a+2d)+3
(12)+(12r)+(12rē) = (12)+(12+d)+(12+2d)+3
12+12r+12rē = 12+12+d+12+2d+3
12+12r+12rē = 39+3d
12r+12rē = 27+3d
Every term can be divided by 3
4r+4rē = 9+d
So we have this system of two equations and 2 unknowns:
12r = 12+d, 4r+4rē = 9+d
Solve the first for d:
12r = 12+d
12r-12 = d
Substitute in
4r+4rē = 9+d
4r+4rē = 9+(12r-12)
4r+4rē = 9+12r-12
4r+4rē = 12r-3
4rē-8r+3 = 0
(2r-1)(2r-3) = 0
2r-1=0; 2r-3=0
2r=1; 2r=3
r=
; r=
Those are the two values.
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To check we must find d
Using r =
12r-12 = d
12
-12 = d
6-12 = d
-6 = d
Arithmetic sequence: 12, 6, 0, ...
Geometric sequence: 12, 6, 3, ...
Sum of 1st three terms of arithmetic sequence = 12+6+0 = 18
Sum of 1st three terms of geometric sequence = 12+6+3 = 21
21 is 3 more than 18
That checks.
Using r =
12r-12 = d
12
-12 = d
18-12 = d
6 = d
Arithmetic sequence: 12, 18, 24, ...
Geometric sequence: 12, 18, 27, ...
Sum of 1st three terms of arithmetic sequence = 12+18+24 = 54
Sum of 1st three terms of geometric sequence = 12+18+27 = 57
57 is 3 more than 54
That checks, too. So it's correct
Edwin