 | Joseph Ray - Algebra - 1848 - 250 pages
...Therefore, *= =— = =•. r—1 r—l Hence, the RULE, FOR FINDING THE SUM OF A GEOMETRICAL SERIES. Multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio less one. EXAMPLES. 1. Find the sum of 10 terms of the progression 2, 6, 18, 54, &,1:. The last term =2x39=2X19683=39366.... | |
 | Jeremiah Day, James Bates Thomson - Algebra - 1848 - 264 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. Obaer. From the above formula, in connexion with the one. iu Art. 368, there may be the same variety... | |
 | Daniel Adams - Arithmetic - 1848 - 324 pages
...required. Hence, RULE. Multiply the larger term by the ratio, and subtract the less term from the product, divide the remainder by the ratio less 1 ; the quotient will be the sum of the series. EXAMPLES FOR PRACTICE. 2. If the extremes be 4 and 131073, and the ratio 8, what is the sum of the... | |
 | Horatio Nelson Robinson - Algebra - 1848 - 354 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 OF EQUATIONS (1) AND (2). 1. Required the sum of 9 terms of the series,... | |
 | Joseph Ray - Algebra - 1848 - 252 pages
...— — ^— = - =•. r — 1 i — l Hence, the RULE, FOR FINDING THE SUM OF A GEOMETRICAL SERIES. Multiply the last term by the ratio, from the product subtract the frst term, and divide the remainder by the ratio less one. EXAMPLES. 1. Find the sum of 10 terms of... | |
 | Nathan Daboll, David Austin Daboll - Arithmetic - 1849 - 260 pages
...last term, (or the extremes,) and the ratio given, to find the sum of the series. RULE. Multiply tke last term by the ratio ; from the product subtract...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... | |
 | Benjamin Naylor - 1850 - 334 pages
...(2) (the first term of the first.) Hence the RULE. product will be the last or greater extreme 2- — multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio less one for the sum of the series, or raise the ratio to a power equal to the number of terms ; subtract... | |
 | Charles Guilford Burnham - 1850 - 350 pages
...find the sum of the series, we have the following l RULES. I. Multiply the last term by the ratio, and from the product subtract the first term, and divide...remainder by the ratio less 1; the quotient will be the answer. II. Divide the difference betiveen the two extremes by the ratio less 1, and add the quotient... | |
 | Horatio Nelson Robinson - Algebra - 1850 - 256 pages
...gives the following rule for the sum of a 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. GENERAL EXAMPLES IN GEOMETRICAL PROGRESSION. 1. What is the ratio of the series 2, 6, 18, 54,... | |
 | John Fair Stoddard - Arithmetic - 1852 - 320 pages
...ART. 197. Given the first term, the last term, and th« ratio, to find the sum of all the terms. RULE. Multiply the last term, by the. ratio ; from the product...first term, and divide the remainder by the ratio di minisked by one. 1 . The first term of a geometrical progression is 4, tha iast term is 62500, and... | |
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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. 
