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
thanks in advance =]
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 $
This one is not an "additive" sequence . . . try ratios.and one for annual salary
$\displaystyle (b)\;\; 42,\!000,\;43,\!680,\; 45,\!427.20,\; 47,\!244.29,\;\hdots$
. . $\displaystyle \begin{array}{ccc}\dfrac{43,\!680}{42,\!000} &=& 1.04 \\ \\[-2mm]
\dfrac{45,\!427.20}{43,\!680} &=& 1.04 \\ \\[-2mm]
\dfrac{47,\!244.29}{45,\!427.20} & \approx & 1.04 \end{array}$
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 $\displaystyle r = 1.04$
. . $\displaystyle \boxed{a_n \;=\;42,\!000(1.04)^{n-1}}$