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The 1st, 4th and 8th terms of an arithmetic sequence, with common difference d, d ≠ 0 , are the first three terms of a geometric sequence, with common ratio r. Given that the 1st term of both sequences is 9 find the value of d and the value of r

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