| 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... | |
| John Mair - Arithmetic - 1772 - 376 pages
...extremes, and common difference, to find the number of terms 5 that is, given I. II. IV. to End III. **RULE. Divide the difference of the extremes by the common difference, and the** quot plus unity is the number of terms, by Theorem IV. EXAMPLE I. A fetting out on a journey, travels... | |
| Mathematics - 1801 - 426 pages
...fast term, the last term, and the common dIfference, to Jind the number of Serins. P.U1.E.* . jpivide **the difference of the extremes by the common Difference, and the quotient,** increased by i, is the number of terms required. . ~" EXAMPLES. * By the last problem,' the difference... | |
| William M. Finlay - Accounting - 1803 - 276 pages
...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... | |
| Thomas Hodson - Arithmetic - 1806 - 492 pages
...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... | |
| 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...EXAMPLES. i. The extremes are 3 and 39, and the common** difCerence 2 ; what is the number of terms \ IE Extremes. Common difference=z)36 Quotient = 18 Add... | |
| Samuel Webber - Mathematics - 1808 - 470 pages
...distance 366 miles. PR0BLEM III. Given thefirst term, the last term, and the common difference^ tofmd **the number of terms. RULE.* Divide the difference...extremes by the common difference, and the quotient,** increased by 1, is the number of terms required. * By the last problem, the difference of the extremes,... | |
| Nicolas Pike - Algebra - 1808 - 468 pages
...PROBLEM PROBLEM III. Given the extranet and the common difference, to fad the number of terms, RutE.* — **Divide the difference of the extremes by the common difference, and the quotient** increased by 1 will be the number of terms required. EXAMPLES. 1 st. The extremes are 3 and 39, and... | |
| Nicolas Pike - Arithmetic - 1809 - 312 pages
...ICOO+IXIOOO 5=500500 Ans. 2PROBLEM III. Giventhtextremtf -and the common difference> to find the numbir **of terms. RULE. Divide the difference of the extremes...common difference, and the quotient increafed by i** wilt be the number of. terms required. EXAMPLES. i. The extremes are 3 and 39, and the common difference... | |
| American - Arithmetic - 1811 - 216 pages
...ftrike in 12 hours ? extremes 1+- 12=13,x (i 12=) 6=78 A. 2; or 1н- 12=13 x 12=150(78 times. A. RULE 2. **Divide the difference of the extremes by the common difference, and the quotient increafed by** one is the number of terms. Ex. if a man gave his youngeft fon 400 S, the next 630, &c. increafing... | |
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