 | John M'Nevin - Arithmetic - 1841 - 300 pages
...of terms given. 2. Multiply the product by the first term, and the result will be the last term. 3. Multiply the last term by the ratio; from the product...term, and divide the remainder by the ratio less 1, for the sum of the series. 1 . If I buy 1 6 cords of wood, and agree to pay 2 cents for the first,... | |
 | Roswell Chamberlain Smith - Arithmetic - 1841 - 324 pages
...when the extremes and ratio are given, to find the sum of the aeries, we have the following RULE. 21. Multiply the last term by the ratio, from the product...term, and divide the remainder by the ratio, less 1 ; the quotient irill be the sum of the series required. 22. If the extremes be 5 and 6,400, and the... | |
 | Roswell Chamberlain Smith - Arithmetic - 1842 - 320 pages
...the extremes and ratio are given, to find the sum of the series, we have the following RULE. 2 1 . 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. 22. If the extremes be 5 and 6,400, and the... | |
 | Roswell Chamberlain Smith - Arithmetic - 1843 - 310 pages
.....- .- >• . • :•••••": •> RULE. _ ........ ' ., • f, g 1'iir1:' ' • • • » J 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. . . , 9. If the extreme! be 5 and 6400, and... | |
 | Roswell Chamberlain Smith - Arithmetic - 1843 - 320 pages
...when the extremes and ratio are given, to find the sum of the series, we have the following RULE. 21. 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. 22. If the extremes be 5 and 6,400, and the ratio... | |
 | Nathan Daboll - Arithmetic - 1843 - 254 pages
...first term, the last term, (or the extremes,) and the ratio given, to find the sum of the series. BULB. Multiply the last term by the ratio ; from the product...term, and divide the remainder by the ratio, less 1, and the quotient will be the sum of all the terms. EXAMPLES. 1 . A man bought 6 yards of cloth, giving... | |
 | James Bates Thomson - Algebra - 1844 - 272 pages
...term in the given series. 373. To find the sum of a geometrical series. Multiply the last term into the ratio, from the product subtract the first term, and divide the remainder by the ratio less one. Obser. From the above formula, in connexion with the one in Art. 368, there may be the same variety... | |
 | James Bates Thomson - Algebra - 1844 - 272 pages
...series. 373. To find the sum of a geometrical series. Multiply the last term into the ratio, from {he product subtract the first term, and divide the remainder by the ratio less one. Obter. From the above formula, in connexion with the one in Art. 368, there may be the same variety... | |
 | Arithmetic - 1845 - 196 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... | |
 | Horatio Nelson Robinson - Algebra - 1846 - 276 pages
...following rule for the sum of a geometrical series: RULE. Multiply the last term by the ratio, and from the , product subtract the first term, and divide the remainder by the ratio less one. EXAMPLES FOR THE APPLICATION OP EQUATIONS (1) AND (2). 1. Required the sum of 9 terms of the series,... | |
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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. 
