If I have 599 matchsticks to make my last two figures what would the answer be if my equation is 4+3(n-1) and 4+3(n-2)
Thank you for the opportunity to help you with your question!
With the information given, it appears that n = the number of match stick that you have
4 + 3(599 - 1) = 4 + 3(598) = 1798
4 + 3(599 - 2) = 4 + 3(597) = 1795
4+3(n-1)=m to the n power and 4=3(n-2)=m to the n power the total of the two equal the 599. It is wanting you to find n
Ok. I see now. I'll work this out for you in a moment.
We are dealing with an arithmetic sequence which follows the formula:
an = a1 + d(n - 1) where a1 is the first term of the sequence, n is the nth term of the sequence, and d is the difference between each term.
We know a1 = 4, and d = 3, and the last two terms add up to 599
Then for the last two terms:
4 + 3(n-1) + 4 + 3(n-2) = 599
8 + 6n - 9 = 599
6n - 1 = 599
6n = 600
n = 100
Therefore, the number of matchsticks in the 100th matchstick square is
a100 = 4 + 3(100 - 1)
= 4 + 3(99) = 4 +297 = 301
and the next to the last is 3 less than that, which is 301 - 3 = 298
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