1. determining formula for sequences.

Im kinda having trouble trying to determine the general forumla for these sequences.

5,4,9,13,22

and one for annual salary

42 000, 43 680, 45 427.20, 47 244.29

2. Originally Posted by ferken
Im kinda having trouble trying to determine the general forumla for these sequences.

5,4,9,13,22

and one for annual salary

42 000, 43 680, 45 427.20, 47 244.29

For the first one, I'm not sure, but it works actually,

Let the pth term be n
so the next term will be n + (p-1)th term

3. To get the general term for the annual salary you just have to take the product of the previous term with 1.04

43 680 = 42 000(1.04)
45 427.20 = 43 680(1.04)
47 244.29 = 45 427.20(1.04)

Hope you will be able to derive the general formula based on the help provided

Originally Posted by ferken
Im kinda having trouble trying to determine the general forumla for these sequences.

5,4,9,13,22

and one for annual salary

42 000, 43 680, 45 427.20, 47 244.29

4. Hello, ferken!

Determine the general forumla for these sequences.

$\displaystyle (a)\;\;5, \:4, \:9, \:13, \:22, \hdots$

If you look, you see that each term is the sum of the preceding two terms.

. . $\displaystyle \begin{array}{ccc}5+4 &=& 9 \\ 4 + 9 &=& 13 \\ 9 + 13 &=& 22 \\ 13 + 22 &=& 35 \\ \vdots && \vdots \end{array}$

Therefore: .$\displaystyle \boxed{a_n \:=\:a_{n-1} + a_{n-2} }\quad\text{with }a_1 = 5,\;a_2 = 4$

and one for annual salary

$\displaystyle (b)\;\; 42,\!000,\;43,\!680,\; 45,\!427.20,\; 47,\!244.29,\;\hdots$
This one is not an "additive" sequence . . . try ratios.

. . $\displaystyle \begin{array}{ccc}\dfrac{43,\!680}{42,\!000} &=& 1.04 \\ \\[-2mm] \dfrac{45,\!427.20}{43,\!680} &=& 1.04 \\ \\[-2mm] \dfrac{47,\!244.29}{45,\!427.20} & \approx & 1.04 \end{array}$

Each term is 1.04 times the preceding term . . .

. . $\displaystyle \boxed{a_n \;=\;1.04a_{n-1}}$

It is a geometric sequence with: first term $\displaystyle a = 42,\!000$ and common ratio $\displaystyle r = 1.04$

. . $\displaystyle \boxed{a_n \;=\;42,\!000(1.04)^{n-1}}$