 | William Frothingham Bradbury - Arithmetic - 1879 - 446 pages
...remainder must be the sum of the given series ; but 4 is the ratio less 1. Hence, Bole. Multiply tJie last term by the ratio, from the product subtract...first term, and divide the remainder by the ratio less one. 106. The extremes are 4 and 2916, and the ratio 3 ; what is the sum of the series? Ans. 4372.... | |
 | Shelton Palmer Sanford - Algebra - 1879 - 348 pages
...can substitute Ir for ar". Hence, we have the following RULE.— Multiply the last term by the ratio, subtract the first term, and divide the remainder by the ratio less 1. EXAMPLES. 2. What is the sum of a geometric series the first term of which is 2, ratio 3, and number... | |
 | Benjamin Greenleaf - Algebra - 1879 - 350 pages
...Substituting the value of arn in (4), ,-£=•:• en Hence the RULE. Multiply the last term Ъу the ratio, subtract the first term, and divide the remainder by the ratio less 1. NOTE. If the last term is not given, it may be found by Case' I.; or, formula (4) may be used instead... | |
 | James Bates Thomson - Algebra - 1880 - 324 pages
...the given series. Substituting Ir for arn, we have -n TT Ir — a FOBMTJLA II. S = Г — I EULE. — Multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio less one. ' For the method of finding the sum of an infinite descending series, see Art. 435. i. Given a... | |
 | William Frothingham Bradbury - 1881 - 404 pages
...the sum of the given series ; but 4 is the ratio less 1. Hence, Rule. Multiply the last term by ihe ratio, from the product subtract the first term, and divide the remainder by the ratio less one. 106. The extremes are 4 and 2916, and the ratio 3 ; what is the sum of the series? Ans. 4372.... | |
 | James Bates Thomson - Algebra - 1884 - 334 pages
...last term in the given series. Substituting Ir for ar", we have FORMULA II. s = r — i RCTLE. — Multiply the last term by the ratio, from the product...subtract the first term, and divide the remainder ly the ratio less one. ' For the method of finding the sum of an infinite descending serifs, see Art.... | |
 | Andrew Jackson Rickoff - Arithmetic - 1886 - 688 pages
...increasing geometrical series, to find the sum of the terms we have the y 4-75. Rule.— Find the last term; multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio — 1. To find the sum of a decreasing geometrical scries, the following is the 4-76. Rule. — Find... | |
 | Andrew Jackson Rickoff - 1888 - 464 pages
...increiis ing geometrical series, to find the sum of the terms we have the 475. Rule.—Find the last term; multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio — 1. To find the sum of a decreasing geometrical scries, the following is the 476. Rule.—Find the... | |
 | William Frothingham Bradbury, Grenville C. Emery - Algebra - 1889 - 428 pages
...Subtracting (1) from (2), rs — s = ¿ r — а Whence, « = — -° . Hence, Ru1e. Multiply tlie last term by the ratio, from the product subtract...first term, and divide the remainder by the ratio less one. 1. Given а = 5, I = 320, and r = 2, to find s. Ir — a 320x2-5 2 — 1 = 635 Aus. 2. Given a... | |
 | William Frothingham Bradbury, Grenville C. Emery - Algebra - 1889 - 444 pages
...(1) from (2), rs — s = lr — a Whence, s= r~a. Hence, T ~~~ i. Bu1e. Multi.ply the last term Ьу the ratio, from the product subtract the first term, and divide the remainder Ьу the ratio less one. 1. Given a — 5, I = 320, and r = 2, to find s. Ir -a, 320X2-5 8 = r = =... | |
| |
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. 
