# How to solve this sequence problem?

$$a_1, a_2, a_3\dots\;$$ 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 $$k.$$

• This is the same as $a_{n+1} = (k+1)a_n$ which is an exponential function. – Ben Dec 29 '18 at 14:23
• 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. – lulu Dec 29 '18 at 14:24
• Yes @lulu But what we do next? – Aether Dec 29 '18 at 14:25
• To get $k$ note that $\frac {a_{1400}}{a_1}=\frac {20000}{20}$ – lulu Dec 29 '18 at 14:27
• @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? – Aether Dec 29 '18 at 14:35

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 $$\frac{1- (1+ k)^n}{k}= 70$$.
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 $$k=\sqrt[1399]{1000}-1\approx0.00494984801948955147$$