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

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

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step-by-step explanation: