Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio, less 1 ; the quotient will be the sum of the series required. A Treatise on Algebra - Page 196by Elias Loomis - 1855 - 316 pagesFull view - About this book
| Zachariah Jess - Arithmetic - 1810 - 222 pages
...is the last term or greater extreme. Multiply the !ast term by the ratio, from the produft subtraft the first term, and divide the remainder by the ratio less one ; the quotient will be the sum of the series. - - . EXAMPLE S. - . j Sold 24 yards of Holland, at га.... | |
| Zachariah Jess - Arithmetic - 1824 - 228 pages
...product is the last term or greater extreme. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less one ; the quotient will be the sum of the series. EXAMPLES. 1 Sold 24 yards of Holland, at 2d. for the... | |
| Zachariah Jess - Arithmetic - 1827 - 226 pages
...term. Then to find the sum of all the terms, multiply the last term by the ratio ; from the product, subtract the first term, and divide the remainder by the ratio, less one ; the quotient will be the sum of all the terms. Or shorter t thus : Involve the ratio to the power... | |
| Thomas Tucker Smiley - 1830 - 188 pages
...term, and that product will be the last term. 3. Multiply the last term by the ratio; from the pro duct subtract the first term, and divide the remainder by the ratio, less 1, for the sum of the series. Questions. What is Geometrical Progression? What is the ratio ? By what... | |
| Roswell Chamberlain Smith - Arithmetic - 1831 - 286 pages
...the series, we have the following easy HULE. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio, less 1 ; the quotient will be the sum of the series required. ft If the extremes be 5 and 6400, and the... | |
| Charles Potts - Arithmetic - 1835 - 202 pages
...the first | term, will give the last term. 2. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less 1, for the sum of the series. EXAMPLES. 1. A labourer wrought 20 days, and received for the first day's... | |
| Roswell Chamberlain Smith - Arithmetic - 1836 - 308 pages
...the Series, we have the following easy RULE. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio, less 1 ; the quotient will be the sum of the series required. 9. If the extremes be 5 and 6400, and the... | |
| A. Turnbull - Arithmetic - 1836 - 368 pages
...extending the series to one more than the given term, and from the last term of the series thus extended, subtract the first term, and divide the remainder by the ratio, less 1. The reason for this rule may be explained thus : * = 8 + 24 + 72 + 216 + 648 + 1944. . If we multiply... | |
| Arithmetic - 1838 - 218 pages
...given, to find the sum of the series. RULE. — Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less I, the quotient will be the sum of the series. The reason of the rule may be shown in the following... | |
| Jason M. Mahan - Arithmetic - 1839 - 312 pages
...less one, and the product will be the last term. Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less one ; the quotient will be the sum of the series. Examples. 123 4 indices. 6 36 216 1296 leading terms.... | |
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