sequences formula

**13373l1t3** · September 18th 2008, 02:45 PM

http://keypress.com/documents/daa1/M...AA_MPYS_01.pdf

WORK SHEET 1.3 Problem #4 a-h

THANK YOU!

**Soroban** · September 18th 2008, 04:05 PM

Hello, 13373l1t3!

They asked for the "recursive" formula.
But if they want, say, the 21st term, we'd better have a "closed" form.

4. Write a recursive formula to generate each sequence.
Then find the indicated term.

$a)\;\;-15, -11, -7, -3, \hdots$ . Find the 10th term.

The sequence increases by 4 at each stage.

The recursive formula is: . $a_n \:=\:a_{n-1} + 4$
(Each term is the preceding term plus 4.)

The closed formula is: . $a_n \:=\:-19 + 4n$

Therefore: . $a_{10} \:=\:-19 + 4(10) \:=\:21$

$b)\;\;1000, 100, 10, 1, \hdots$ . Find the 12th term.

Each term is one-tenth of the preceding term.

The formula is: . $a_n \:=\:\frac{10,000}{10^n}$

Therefore: . $a_{12} \:=\:\frac{10,000}{10^{12} }\:=\:10^{-8} \:=\:0.00000001$

$c)\;\;17.25, 14.94, 12.63, 10.32, \hdots$ . Find the 15th term.

The terms decrease by 2.31

The formula is: . $a_n \:=\:19.56 - 2.31n$

Therefore: . $a_{15} \:=\:19.56 - 2.31(15) \:=\:-15.09$

$d)\;\;0.3, -0.03, 0.003, -0.0003,\hdots$ . Find the 8th term.

Each term is $-0.1$ times the preceding term.

The formula is: . $a_n \:=\:0.3(-0.1)^{n-1}$

Therefore: . $a_8 \;=\;0.3(-0.1)^7 \:=\:0.00000003$

$e)\;\;0, \frac{1}{6},\:\frac{1}{3},\:\frac{1}{2},\hdots$ . Find the 21st term.

This is a sneaky one!
Here is my thought process . . .

The numerators seems to be all 1's.
Then how come the first term has a numerator of 0 ?

Then it hit me . . . the numerators are: . $0,1,2,3,4,\hdots$
. . and the denominators are all 6's !!

Get it? .We have: . $\frac{0}{6},\:\frac{1}{6},\:\frac{2}{6},\:\frac{3} {6},\:\frac{4}{6},\hdots$

. . and that is why the sequence begins: . $0,\:\frac{1}{6},\:\frac{1}{3},\:\frac{1}{2},\:\hdo ts$

The formula is: . $a_n \;=\;\frac{n-1}{6}$

Therefore: . $a_{21} \;=\;\frac{21-1}{6} \:=\:\frac{20}{6} \:=\:\frac{10}{3}$