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$a_1, a_2, a_3\dots\;$ are numbers in a sequence with the condition $$a_{n+1}= a_n + ka_n$$ where k is a constant.
If the first number is $20,$ the last is $20000$ and total numbers are $1400.$ Find $k.$

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  • $\begingroup$ This is the same as $a_{n+1} = (k+1)a_n$ which is an exponential function. $\endgroup$
    – Ben
    Dec 29, 2018 at 14:23
  • $\begingroup$ It appears that you are saying $a_{n+1}=(1+k)a_n$, no? So then the final sum is essentially a (finite) geometric series. $\endgroup$
    – lulu
    Dec 29, 2018 at 14:24
  • $\begingroup$ Yes @lulu But what we do next? $\endgroup$
    – Aether
    Dec 29, 2018 at 14:25
  • $\begingroup$ To get $k$ note that $\frac {a_{1400}}{a_1}=\frac {20000}{20}$ $\endgroup$
    – lulu
    Dec 29, 2018 at 14:27
  • $\begingroup$ @lulu okay. But I m doing this for music theory. And i cannot dig into sequence and series right now. My basics aren't good in that topic. Would you please find k for me? $\endgroup$
    – Aether
    Dec 29, 2018 at 14:35

2 Answers 2

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This is, as others said, a geometric sequence with common ratio 1+ k. $a_n= a_1(1+ k)^{n-1}$ and the sum, to k= n, is $a_1\frac{1- (1+ k)^n}{k}$. "The first number is 20". So $a_1= 20$. "The last number is 20000". So $a_n= 20(1+ k)^{n-1}= 20000$. $(1+ k)^{n-1}= 1000$. "The sum is 14000". That's impossible. The sum is 20000 plus other positive numbers so must be larger than 20000. I am going to try "The last number is 2000" rather than 20000. Then we have $(1+ k)^{n-1}= 100$ and $20\frac{1- (1+ k)^n}{k}= 1400$ or $\frac{1- (1+ k)^n}{k}= 70$.

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The first number is $a_1=20$, the last number is $a_{1400}=20000$. The definition tells you that $$ a_{1400}=20(1+k)^{1399} $$ so $$ (1+k)^{1399}=1000 $$ and therefore $$ k=\sqrt[1399]{1000}-1\approx0.00494984801948955147 $$

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