 | Roswell Chamberlain Smith - Arithmetic - 1831 - 286 pages
...when the extremes and ratio are given, to find the sum of the series, we have the following easy HULE. Multiply the last term by the ratio, from the product...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
...lessll than the number of terms given, and multiply it by the first | term, will give the last term. 2. Multiply the last term by the ratio, from the product...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... | |
 | Stephen Pike - Arithmetic - 1835 - 210 pages
...terms given, which, being multiplied by the first term, will give the last term, or greater extreme. 2. Multiply the last term by the ratio, from the product...subtract the first term, and divide the remainder by ratio less one for the sum of the series. . EXAMPLES. 1. A thresher wrought 20 days, and received for... | |
 | Roswell Chamberlain Smith - Arithmetic - 1836 - 308 pages
...When the Extremes and Ratio are ¡riven, to ßnd thf Sum of the Series, we have the following easy RULE. Multiply the last term by the ratio, from the...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
...last ? Ans. $171798691,84. CASE 2. The extremes and ratio being given, to find the sum of the series. RULE. — Multiply the last term by the ratio, from...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
...of the ratio as is indicated by the number of terms less one, and the product will be the last term. Multiply the last term by the ratio, from the product...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.... | |
 | Roswell Chamberlain Smith - 1839 - 308 pages
...When the Extremes and Ratio are given, to find tlte Sum of the Series, we have the following easy «rf RULE. Multiply the last term by the ratio, from the...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... | |
 | Nathan Daboll - 1839 - 220 pages
...The first term, the last term, (or the extremes,) and the ratio given, to find the sum of the series. RULE. Multiply the last term by the ratio ; from the...first term, and divide the remainder by the ratio, loss 1, and the quotient will be the sum of all the terms. EXAMPLES. 1. A man bought 6 yards of cloth,... | |
 | John H. Harney - Algebra - 1840 - 298 pages
...— a (r — l)s = ar5 — a ar5 — a Multiply the last term of the series by the ratio ; from that product subtract the first term, and divide the remainder by the ratio, less one. The result will be equal to the sum of the series. 10 Find the sum of 5 terms of the series 60, 120,... | |
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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. 
