Given the first term, last term, and common difference, to find the number of terms. RULE. — Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms. The parallel arithmetic - Page 106by W H. Wingate - 1865Full view - About this book
| Charles Hutton - Arithmetic - 1766 - 191 pages
...and the debt is 135/. 4*. P«QPROBLEM VI. Given the extremes a'nd the common difference, to find 1 . **The number of terms. RULE. Divide the difference of the extremes by the common difference,** add i to the quotient, and the fum }vill be the number of terms. 2. The fum of the feries. Having found... | |
| Anthony Birks, John Birks - Arithmetic - 1766 - 642 pages
...difference. Which added to each day's journey, PROPOSITION V. The two extremes, and the common excefs **given, to find the number of terms. RULE. Divide the difference of the** two extremes by the common excefs, the quotient plus unity is the number of terms. ¿k 2 8. A gives... | |
| John Thomas Hope - Arithmetic - 1790 - 387 pages
...and the whole diftancc 155 coli from Calcutta. PROPOSITION V. The two extremes and the common excefs **given to find the number of terms. RULE. Divide the difference of the** two extremes by the common exceli, the quotient plus unity is the number of terms. (l) A man going... | |
| Thomas Peacock - Arithmetic - 1791 - 283 pages
...find the common difference ? Anf. iiy, , # PROBLEM III. The firft term, the laft term, and the common **difference being given, to find the number of terms. RULE. Divide the difference of the** extreme» by the common difference, and add i to the quotient for the anfwer, E XA MP LE S. 1. The... | |
| Nicolas Pike - Arithmetic - 1802 - 352 pages
....iboo+i X 1000 " / =500500 Anfwer. 2 PROBLEM 3. Given the extremis and the common dtffsr~ fnce, to Jind **the number of terms. RULE. — Divide the difference of the extremes by the common difference,** and the quotient increafed by i will be the number of terms required. EXAMPLES. I. The extremes are... | |
| William M. Finlay - Accounting - 1803 - 276 pages
...18, then 36-7-18=2 common dif. PROBLEM V. Given the first term, the last term, and common difference, **to find the number of terms. RULE.— •Divide the...difference of the extremes by the common difference—** Jhe quotient + lj "»ñu be the number of terms required. EXAMPLE. Given the first, 7 the last, S l... | |
| Paul Deighan - Arithmetic - 1804 - 504 pages
...10+15+20, &c. the whole debt» 275!. Proportion 3. When the two extremes and common difference are gjven, **to find the number of terms. Rule. Divide the difference of the** :wo extremes by the common difference or excels ; add unity or i to the quotient, and thcrfani will... | |
| Thomas Hodson - Arithmetic - 1806 - 492 pages
...quotient 2 is the common difference. PROBLEM II. Having the two extremes and the common difference, **to find the number of terms. Rule. Divide the difference of the extremes by the** com- VOL. I. li own mon difference, and i added to the quotient will be the number of terms. Example... | |
| Charles Vyse - Arithmetic - 1806 - 342 pages
...and what must each Payment be ? PROPOSITION III. When the two Extremes and the common Difference are **given, to find the Number of Terms. RULE. Divide the Difference of the** two Extremes by the common Excess or Difference ; add Unity or 1 to the Quotient, and the Sum will... | |
| Nicolas Pike - Arithmetic - 1807 - 352 pages
...? 1000+ i X 1000 =500500 Anf<wer. 2 PROBLEM 3 Given the extremes and the common difference, tojlnd **the number of terms. RULE — Divide the difference of the extremes by the common difference,** and the quotient increafed by i will be the number of terms required. EXAMPLES. i. The extremes are... | |
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