 | 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... | |
 | 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...EXAMPLES. I. The extremes are 3 and 39, and the common difference 2 ; what is the number of terms ? t Ч * • ' «* 3' £ Extremes. ^P ' • Common differences:... | |
 | 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... | |
 | 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,... | |
 | 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... | |
 | Jeremiah Joyce - Arithmetic - 1812 - 276 pages
...III. The extreme terms a and z, and common difference d being given, to find the number of terms n. RULE. Divide the difference of the extremes by the common difference, and the quotient increased by unity is the number sought : or + i ~ n. Ex. 1. When the extremes are 4 and 106, and the... | |
 | Samuel Webber - Arithmetic - 1812 - 266 pages
...distance 366 miles. PROBLEM 3. Given the first term, the last term, and the common difference^ to find the number of terms. % RULE.* Divide the difference of the extremes by the common dlf* By the last problem, the difference of the extremes, divided by the number of terms less 1, gives... | |
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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. 
