SOLUTION: Hi could you please guide me through this? Thank you so much, and for your time! Given a sequence is arithmetic find the 4 terms between 6 and 121. Also: Given the sequ

Question 634686: Hi could you please guide me through this? Thank you so much, and for your time!
Given a sequence is arithmetic find the 4 terms between 6 and 121.

Also:
Given the sequence is geometric find the 3 terms between 6 and 3.9366.
Thank you so much with these problems! Thank you a lot!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1

Let d = common difference

So

6+d is the next term (after 6)

6+d+d = 6+2d is the next term (after the previous term above)

6+d+d+d = 6+3d is the next term (after the previous term above)

6+d+d+d+d = 6+4d is the next term (after the previous term above)

6+d+d+d+d+d = 6+5d is the next term (after the previous term above)

So we have the sequence...

6, 6+d, 6+2d, 6+3d, 6+4d, 6+5d

We have 6 terms and 4 terms are sandwiched between the terms "6" and "6+5d"

So the term 6+5d must be the term 121. So 6+5d = 121

Solve this for d. Then use this solution to find the missing 4 terms.

=======================================================
# 2

Let r = common ratio

So

We start off with 6

Then we move to 6*r or 6r to get the next term

Then we multiply by r again to get 6r^2 as the next term

Then we repeat to get 6r^3

Repeat again to get 6r^4

So we have the 5 terms

6, 6r, 6r^2, 6r^3, 6r^4

There are 3 terms in between the first term 6 and the last term 6r^4. The last term given to us is 3.9366

So

6r^4 = 3.9366

Solve this equation for r. Then use it to find the missing terms.

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