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Sagot :
Answer:
d=1, r=4/3
Step-by-step explanation:
For aritmetic sequence
a1=9
a1+3d=a4 9+3d=a4.
a1+7d=a8 9+7d=a8.
For geometric sequence
b1*r=b2
b2=a4, b1=a1=9
9*r=a4
b3= b1*r^2
b3=a8
a8=9*r^2
So 9+3d=a4
9+3d=9r
9+7d= a8=9r^2
So we have a system of equations
9+3d=9r
9+7d=9r^2
Multiply 9+3d=9r by -7 It is equal to -63-21d=-63r
And multiply 9+7d=9r^2 by 3. It is equal to 27+21d=27r^2
Then add them
-63-21d+ (27+21d)= -63r+27r^2
-36+63r-27r^2=0 /9
-4+7r-3r^2=0
3r^2-7r+4=0
r=1 r=4/3
If r=1 9+3d=9
d=0 (It is not right, d isn't equal to 0)
If r=4/3
9+3d= 9*4/3
9+3d=12
d=1
The value of the common difference, d is 9
The value of the common ratio r is 2
Arithmetic and Geometric Sequences
The nth term of an arithmetic sequence is:
[tex]T_n=a+(n-1)d[/tex]
The first, fourth, and eighth terms are therefore:
a, a + 3d, a + 7d
The nth term of a geometric sequence is:
[tex]T_n=ar^{n-1}[/tex]
The first three terms of the arithmetic sequence are:
[tex]a, ar^2, ar^3[/tex]
For both sequences, the first term is 9
That is, a = 9
[tex]a+3d=ar^2\\\\9+3d=9r^2..............(1)\\\\a+7d=ar^3\\\\9+7d=9r^3..........(2)[/tex]
Make d the subject of the formula in equation (1) and substitute to equation (2)
[tex]d=\frac{9r^2-9}{3} \\\\d=3r^2-3[/tex]
Substitute equation (3) into equation (2)
[tex]9+7(3r^2-3)=9r^3\\\\9+21r^2-21=9r^3\\\\21r^2-9r^3=12\\\\7r^2-3r^3=4\\\\r=1, \frac{-2}{3}, 2[/tex]
Substitute r = 2 into (1)
[tex]9+3d=9(2^2)\\\\d = 9[/tex]
The value of the common difference, d is 9
The value of the common ratio r is 2
Learn more on sequences here: https://brainly.com/question/6561461
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