Question

Find the 100th and the nth term for each of the following sequences. a. 1,3,5,7 b. 50, 90, 130 c. 1.3.9. ... d. 9.93.95.97 e. 142+7.229.142+8.229.142 +9.229

Answer 1

a) the 100th term is 199 and the nth term is 2n-1.

b) the 100th term is 4010 and the nth term is 40n + 10

c) the 100th term is 60.7 and the nth term is .6n +.7

d) The 100th term is 11.91 and the nth term is 9.91 + 0.02n

e) The 100th term is 142 +106.229 and the nth term is 142 + (6.229+n)

Explanation:

The formula to find the nth term of a sequence is:

`a_(n)=a_(1)+d(n-1)`

Where
`a_(n)`is the nth term,
`a_(1)`is the first one and d is the difference between each term.

So now we're going to apply this to our sequences

a) 1, 3, 5, 7...

First term = 1

Difference = 2

nth term =
`1+2(n-1)` = 1 + 2n - 2 = 2n - 1

100th term= 2(100) - 1 = 200 - 1 = 199

b) 50, 90, 130...

First term = 50

Difference = 40

nth term = 50 + 40(n - 1) = 50 +40n - 40 = 40n + 10

100th term = 40(100) + 10 = 4000 + 10 = 4010

c) 1.3, 1.9...

First term = 1.3

Difference = .6

nth term =
`1.3 + .6(n-1)\\1.3 +.6n-.6\\.6n+.7`

100th term = .6(100) +.7 = 60 + .7 = 60.7

d) 9.93, 9.95, 9.97...

First term = 9.93

Difference = .02

nth term =
`9.93 + .02(n-1)\\9.93+.02n-.02\\9.91 +.02n`

100th term = 9.91+.02(100) = 9.91 + 2 = 11.91

e) 142+7.229, 142+8.229...

First term = 142 +7.229

Difference = .02

nth term =
`(142 + 7.229) +1(n-1)\\(142+7.229)+n-1\\142 + (7.229+n-1)\\142 + (6.229+n)`

100th term= 142 + (6.229 + 100) = 142 + 106.229