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.
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# 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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