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Thanks for the pointers! Here's updated and edited question I'm trying to find the number of days it takes to reach 982 miles when you start traveling at 18 miles/day and decrease your speed by 2% each day. Here's the recursive formula I've put together: $$a_n={18,n=1}$$ and $$a_n=\{a_n-_1-.02*a_n-_1 , n>1\}$$

Therefore the series used to determine distance traveled is: $$\sum_{n=1} ((a_n-_1)) -(.02*a_n-_1))=982$$

How do I use this with to find the day at which the 982 mile-mark is reached? Thanks!

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  • $\begingroup$ Its not clear what your asking here. Can you clean up the notation a bit or add some more explanation of what the problem is? $\endgroup$ – Spencer Apr 28 '15 at 22:13
  • $\begingroup$ Here is a through guide for formatting mathematics on this site. Please use it to make your expressions readable. For the question itself: your sequence is of a special kind that has a name: a geometric sequence. Do you know any formulas concerning such sequences that you think might be relevant? $\endgroup$ – Arthur Apr 28 '15 at 22:17
  • $\begingroup$ yes, a/(1-r) and (a_n+_1)/(a_n) $\endgroup$ – jackson Apr 28 '15 at 23:23
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Let $r=2\%=\frac{2}{100}$
First day: $$a_{0}=18$$Second day: $$a_{1}=18-18r=a_{0}-a_{0}r=a_{0}(1-r)$$Third day: $$a_{2}=a_{1}-a_{1}r=a_{1}(1-r)=a_{0}(1-r)^{2}$$ Therefore on n-th day: $$a_{n}=a_{0}(1-r)^{n}$$ So we have: $$\sum_{n=0}^{N} a_{0}(1-r)^{n}=\sum_{n=0}^{N}a_{0}k^{n}$$ This is called geometric series. As $N$ goes to infinity, the absolute value of $k$ must be less than one for the series to converge i.e. $|k|<1$;
$$k=1-\frac{2}{100}=\frac{98}{100}<1$$ so condition is satisfied. It's convenient for this problem to check what happens when $N\rightarrow\infty$. $$\sum_{n=0}^{\infty}a_{0}k^{n}=\frac{a_{0}}{1-k}=\frac{18}{1-\frac{98}{100}}=900$$

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