Question

An arithmetic sequence has 5 and 13 as its first two terms respectively
a. Write down, in terms of n, an expression for the nth term a(sub root)n
b. Find the number of terms of the sequence which are less than 400

Answer 1

The so-called common difference is d=13-5=8. The nth terms is given by the formula a(n)=a1+(n-1)d. Where a1 is the first term. filling in for a1 and d gives a(n)=5+(n-1)*8.

Answer 2

For part b, do mean how many terms are in the sequence that are less than or equal to 400?

Answer 3

I think so. I wrote it in exact terms it just confused me

Answer 4

The 51st term is 405, the 50th term is 397. If you solve 400<=5+(n-1)*8 for n, you get 50.375. But n must be an integer, i.e., n=50 terms.

Answer 5

Is there an equation for solving that?

Answer 6

The nth term a(n) can be found by adding to the first term a1 the product (n-1)*d, where d is the common difference as described above, and n is the number of terms, thus: a(n)=a1+(n-1)*d.

Answer 7

For example, you were given 5, 13, . . . It is obvious that the third term would be 13+8=21, but to use the formula that I just gave you, we could also get 21 by finding a(3)=5+(3-1)*8=5+2*8=5+16=21. You usually would not use that formula to find an adjacent term, but if you want to find say the 101st term it can be very useful. For example, a(101)=5+(101-1)*8=5+100*8=5+800=805.

Answer 8

ok and to do it backwords such as
400= 5+ (n-1)8 would help me find the number of terms right

Answer 9

Correct. That was the method for solution to your part B.

Answer 10

Thank you