40]
SECTION VIII—ARITHMETICAL PROGRESSION
85
As many times as is necessary to multiply 60 to make 100, so many times must these terms be multiplied to make the true series.
\ | 1 | 60 |
\ | 2⁄3 | 40. |
The total, 12⁄3, times 60 makes 100.
Multiply by 12⁄3
23 | it | becomes | 381⁄3 | |
17 | " | " | 291⁄6 | |
12 | " | " | 20 | |
61⁄2 | " | " | 12⁄3 | |
Total | 60 | " | " | 100. |
This problem has been fully explained in the Introduction (page 12) and needs no further comment here.
In this problem and in Problem 64 our author writes his progression as a descending series, and so the smallest term is with him the last term.