d = 5 - 1 = 4 , d = 9 - 5 = 4
d = 4
an = a1 + (n - 1)d
a1 = 1 and d = 4
an = 1 + (n - 1)4
an = 1 + 4n - 4 = 4n - 3
an = 4n - 3
a = first term
l = last term
a = 1, l = a30 = 4(30) - 3 = 120 - 3 = 117
Find the sum of the first 30 terms of the arithmetic sequence 1, 5, 9,...
| First find the nth term. | |
| Solve | Step |
|
d = 5 - 1 = 4 , d = 9 - 5 = 4 |
Find
the common difference d = 4 |
|
an = a1 + (n - 1)d |
Use the formula for arithmetic. |
|
a1 = 1 and d = 4 |
a1 = 1 is the first term of the sequence. |
|
an = 1 + (n - 1)4 |
Plug in the values... |
|
an = 1 + 4n - 4 = 4n - 3 |
... and simplify. |
|
an = 4n - 3 |
nth term |
| Use the formula for the sum of an arithmetic sequence. | |
|
|
n = number of terms a = first term l = last term |
|
a = 1, l = a30 = 4(30) - 3 = 120 - 3 = 117 |
n = 30 |
|
|
Plug in the values... and simplify. |
| Spot Check |
|
a1 = 4(1) - 3 = 1 |
| a3 = 4(3) - 3 = 12 - 3 = 9 |