 | Charles Hutton - Arithmetic - 1766 - 214 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 - 446 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 of the extremes,... | |
 | Nicolas Pike - Arithmetic - 1802 - 350 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...extremes by the common difference, and the quotient increafed by i will be the number of terms required. EXAMPLES. I. The extremes are 3 and 39, and the... | |
 | William M. Finlay - Accounting - 1803 - 272 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 - 502 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 - 370 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...extremes by the common difference, and the quotient increafed by i will be the number of terms required. EXAMPLES. i. The extremes are 3 and 39, and the... | |
 | Samuel Webber - Mathematics - 1808 - 466 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...common difference, and the quotient, increased by 1, is the number of terms required. * By the last problem, the difference of the extremes, divided... | |
 | Nicolas Pike - Algebra - 1808 - 470 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 the common difference... | |
 | 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 by the 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... | |
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Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms. 
