Question

Write the sum using summation notation, assuming the suggested pattern continues.

-9 - 4 + 1 + 6 + ... + 66

Answer 1

@precal

Answer 2

This is an arithmetic series.
Can you find the common difference?

Answer 3

i dont know how to approach the question at all could you walk me through it please?

Answer 4

You should first try to recognize any pattern in the numbers.
Looks at the first two terms (or numbers).
What can be done to the first term to get the second term? By that I mean what can be added or subtracted or divided or multiplied to get the second term from the first term? Then if you repeat the same operation will you get the third term from the second term?

Answer 5

its adding 5 everytime

Answer 6

When I look at -9 and -4 I know that if I add 5 to the first term I will get the second term.
See if the pattern continues. Add 5 to the second term and see if you get the third term. -4 + 5 = 1 which is the third term. Add 5 to the third term and you get the fourth term.

Answer 7

yeah i saw that pattern what do i do next?

Answer 8

correct. If you keep adding 5 you get the next term. Such a pattern is called an arithmetic progression or arithmetic sequence. Arithmetic sequence uses a few terms such as common difference, which in this case is 5. The first term is -9. There is a formula for finding the nth term of an arithmetic sequence. It is:
\(\large a_n = a_1+(n-1)*d = -9 + (n-1)*5 = -9 + 5n - 5 = 5n - 14\)
\(\large a_n = 5n - 14\)

Answer 9

You can try putting n = 1 in the above formula and you will get the first term.
Put n = 2 to get the second term.
n = 3 to get the third term.
You can make sure the formula is correct.

Answer 10

Can you use that formula to find how many terms are there in this series which starts at -9 and ends at 66?

Answer 11

15?

Answer 12

\(\large a_n = 5n - 14 \)

The last term is 66. Set \(\large a_n = 66\) and solve for 'n'.
66 = 5n - 14
66+14 = 5n
5n = 80
n = 80/5 = 16
The last term is the 16th term. That means there are 16 terms in this arithmetic series.