Question

arithmetic sequences

Answer 1

what about them?

Answer 2

could you please help me figure out how to solve them?

Answer 3

if i use the formula an=a1+(n-1)d

Answer 4

To find any term of an arithmetic sequence:

a(sub n) = a1 + (n - 1)d

where a1 is the first term of the sequence,
d is the common difference, n is the number of the term to find.

Answer 5

can we do an example

Answer 6

sure
do you have examples you'd like to run thru, or want me to make up one?

Answer 7

5,8,11,14

Answer 8

find d. the common difference ... so 8-5, 11-8, and 14 - 11 all equal ?

Answer 9

3

Answer 10

3.
our first term a (sub 1) is given as 5
so, a(sub n) = 5 + (nth place -1)3
if you wanted the 12th number in the sequence
a(12) = 5 + (12 - 1)*3 = 5 + 11*3 = 38

Answer 11

I'm sorry, i have no clue what you mean

Answer 12

the equation is always the same "an=a1+(n-1)d" \[a _{n} = a _{1} + (n - 1)d\]

Answer 13

n is the number in the sequence, in your example
5, n = 1
8, n=2
11, n=3 ... n is the position in the sequence

d is the difference between the numbers , in this case 3

Answer 14

so, if you needed the 20th term n = 20
\[a _{20} = a _{1} + (20 - 1)d \]
d = 3
\[a _{20} = a _{1} (20 - 1)3\]
and the first term is 5
\[a _{20} = 5 + (20 - 1)3\]
so the 20th term in the sequence would be 72

Answer 15

i get it now, thank you for you help :)

Answer 16

no worries

Answer 17

how about arithmetic means between two given terms in an arithmetic sequence

Answer 18

in your example (5, 8, 11, 14) the difference (d) is 3
\[a _{1} + a _{last} = 5 + 14 = 19\]
the 4th term is 14 ....(4 - 1 = 3)
3d = 3*3=12
so the arithmetic means between them are
5+3, and 5 +2(3), so 8 and 11 , the terms between the 1st and last are your arithmetic means

Answer 19

what if the problem is 7, , , ,31

Answer 20

typically you find those if you are not given them
if I say the sequence is 1, _, _, _, 5 .. you would figure out it from there,
here the d = 1
and the means would be 2,3,4

Answer 21

would i use the formula like last time

Answer 22

ah .. sorry didn't see your post.
find d
\[a_{1} = 7 , a _{5} = 31\]
5 - 1 (part of the formula above) = 4
7 + 4d = 31
4d = 31 - 7 = 24
d = 24/4 = 6
sequence would be 7, 7+6, 7+(2*6), 7+(3*6), 7+(4*6), ...

Answer 23

7, 13, 19, 25, 31, ...
and the means would be , 13, 19, 25,

Answer 24

but how? i can't figure it out??

Answer 25

in your example 7, , , , 31.
\[n _{1} is 7 \]
\[n _{5} is 31\]
so the sequence is
7 + (0)*d, 7 + (1)*d , 7 + (2)d , 7 + (3)*d , 7 + (4)*d , ...
1st term is ... 7 = 7 + (0)*d
5th term is ... 31 = 7 + (4)*d

Answer 26

find d by solving
7 + 4d = 31 4d = 31 - 7 = 24
d = 24/4 = 6

Answer 27

sequence would be 7, 7+6, 7+(2*6), 7+(3*6), 7+(4*6), ...

Answer 28

I'm sorry i still don't get it

Answer 29

okay, .. what part?
d?
n ?
a?
how I found them?

Answer 30

An arithmetic mean is the term between any two terms of an arithmetic sequence. It is simply the average (mean) of the given terms.