 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 - Side 196av Elias Loomis - 1855 - 316 siderUten tilgangsbegrensning - Om denne boken
 | Zachariah Jess - 1810 - 222 sider
...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 - 1824 - 228 sider
...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 - 1827 - 226 sider
...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 sider
...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 - 1831 - 286 sider
...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 - 1835 - 202 sider
...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 - 1836 - 308 sider
...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 - 1836 - 368 sider
...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... | |
 | 1838 - 218 sider
...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 - 1839 - 312 sider
...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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