five sequence Qs: repeating fractions as rationals, etc.

ladventchildp

New member
Joined
Apr 14, 2007
Messages
2
few questions im not getting can someone help me out

First question: Write an expression for the apparent nth term of the sequence as a function of n: -1, 2, 7, 13, 23, …

I keep getting n^2-2 but that doesnt work with the 4th term

Second Question: Find a formula for nth term of the arithmetic sequence as a function of n, where a5=190 and a10=115

For this, I did a10=a5+5d and then d=-15 and then i believe the formula to find a1 is an=a1+(n-1)(d) but when i find a1 and then the formula an=dn+c it gives me two figures depending on what number i use to find a1, if i use a10 i get a different numer and a15 i get 250-15n=an

Third question: Use the concepts of geometric sequences to write 1.3245 (i.e., 1.324545454545…) as rational number in the form a/b, where a and b are integers.

Fourth question: Let a1=0 a2=1 and ak + 1 = ak + ak - 1 for k ³ 2. Write the first 8 terms of this sequence.

Fifth question: How much money will I have been paid over my 30 year career if my starting salary is $40,000, and I receive a $1,000 salary raise for each year I work?
 
Re: hw question

Hello, ladventchildp!

I don't see a pattern for the first one . . . Is there a typo?


2) Find a formula for \(\displaystyle n^{\text{th}}\) term of the arithmetic sequence as a function of \(\displaystyle n,\)
where: \(\displaystyle a_5\,=\,190\) and \(\displaystyle a_{10}\,=\,115\)

Your general formula is correct: \(\displaystyle \:a_n\:=\:a_1\,+\,(n-1)d\)

So we have: \(\displaystyle \:\begin{array}{cccccc}a_5 & = & a_1\,+\,4d & = & 190 & \:(1) \\ a_{10} & = & a_1\,+\,9d & = & 115 & \:(2)\end{array}\)

Subtract (1) from (2): \(\displaystyle \:5d\,=\,-75\;\;\Rightarrow\;\;d\,=\,-15\)

Substitute into (1): \(\displaystyle \:a_1\,+\,4(-15)\:=\:190\;\;\Rightarrow\;\;a_1\,=\,250\)

Therefore: \(\displaystyle \:a_n\:=\:250\,-\,15(n\,-\,1)\)



3) Use the concepts of geometric sequences to write \(\displaystyle 1.\overline{3245}\)
as rational number in the form \(\displaystyle \frac{a}{b}\) where \(\displaystyle a\) and \(\displaystyle b\) are integers.

We have: \(\displaystyle \:N\:=\:1.3245324532453245\,.\,.\,.\)

\(\displaystyle N \:=\:1\,+\,0.3245\,+\,0.00003245\,+\,0.000000003245\,+\,.\,.\,.\)

\(\displaystyle N \:=\:1\,+\,\frac{3245}{10^4}\,+\,\frac{3245}{10^9}\,+\,\frac{3245}{10^{12}}\,+\,.\,.\,.\)

\(\displaystyle N \:=\:1\,+\,\frac{3245}{10^4}\cdot\underbrace{\left(1\,+\,\frac{1}{10^4}\,+\,\frac{1}{10^8}\,+\,\frac{1}{10^{12}}\,+\,.\,.\,.\right)}_{\text{a geometric series}}\)

We have a geometric series with first term \(\displaystyle a_1\,=\,1\) and common ratio \(\displaystyle r\,=\,\frac{1}{10^4}\)
. . Its sum is: \(\displaystyle \:\frac{1}{1\,-\,\frac{1}{10^4}} \:=\:\frac{10^4}{9999}\)

Therefore: \(\displaystyle \:N\;=\;1\,+\,\frac{3245}{10^4}\left(\frac{10^4}{9999}\right) \:=\:1\,+\,\frac{3245}{9999} \:=\:\frac{13,244}{9,999}\)



4) Let \(\displaystyle a_1\,=\,0,\;a_2\,=\,1\) and \(\displaystyle a_{k+1}\:=\:a_k\,+\,a_{k-1}\) for \(\displaystyle k\,\geq\,2\)
Write the first 8 terms of this sequence.

That general equal says: each term is the sum of the preceding two terms.

Therefore, the sequence is: \(\displaystyle \:0,\,1,\,1,\,2,\,3,\,5,\,8,\,13,\,.\,.\,.\)

. . This is the famous Fibonacci Sequence.



5) How much money will I have been paid over my 30-year career
if my starting salary is $40,000, and I receive a $1,000 salary raise for each year I work?

This is an arithmetic series with first term \(\displaystyle a_1\,=\,40,000\) and common difference \(\displaystyle d\,=\,1000\)

The sum of the first \(\displaystyle n\) terms is: \(\displaystyle \:S_n\:=\:\frac{n}{2}[2\cdot a_1\,+\,(n-1)d]\)

So we have: \(\displaystyle \: S_{30}\;=\;\frac{30}{2}[2(40,000)\,+\,29(1000)] \;=\;\$1,635,000\)

 
ladventchildp said:
Write an expression for the apparent nth term of the sequence as a function of n: -1, 2, 7, 13, 23, …
I keep getting n^2-2 but that doesnt work with the 4th term
Your n^2-2 is correct for every odd n

For evens: n^2 - n/2 - 1 or (2n^2 - n - 2) / 2
 
Top