as before. The rule, then, is: To sum any number of terms of an arithmetical progression, multiply the sum of the first and last terms by half the number of terms. For example, what are 99 terms of the series 1, 2, 3, &c.? The 99th term is 99, and the sum is
| (99 + 1) | 99 | , or | 100 × 99 | , or 4950. |
|---|---|---|---|---|
| 2 | 2 |
The sum of 50 terms of the series
| 1 | , | 2 | , | 1 , | 4 | , | 5 | , 2 , &c. is | ( | 1 | + | 50 | ) | 50 | , |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 3 | 3 | 3 | 3 | 3 | 3 | 2 |
or 17 × 25, or 425.
- The first term being given, and also the common difference and number of terms, the last term may be found by adding to the first term the common difference multiplied by one less than the number of terms. For it is evident that the second term differs from the first by the common difference, the third term by twice, the fourth term by three times the common difference; and so on. Or, the passage from the first to the nth term is made by n-1 steps, at each of which the common difference is added.
EXERCISES.
| Given. | To find. | ||||
|---|---|---|---|---|---|
| Series. | No. of terms. | Last term. | Sum. | ||
| 4, | 6½, | 9, &c. | 33 | 84 | 1452 |
| 1, | 3, | 5, &c. | 28 | 55 | 784 |
| 2, | 20, | 38, &c. | 100,000 | 1799984 | 89999300000 |
- The sum being given, the number of terms, and