 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.  Stoddard's Practical Arithmetic - Page 233by John Fair Stoddard - 1852 - 299 pagesFull view - About this book
 | 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... | |
 | 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 = =... | |
 | Joseph Ray - Algebra - 1894 - 252 pages
...may the common ratio in any geometrical series be found ? TO FIND THE SUM OF A GEOMETRICAL SERIES, Rule. — Multiply the last term by the ratio, from...first term, and divide the remainder by the ratio less one. 1. Find the sum of 10 terms of the progression 2, 6, 18, 54, etc. The last term =2X39 =2... | |
 | William Frothingham Bradbury - Arithmetic - 1895 - 398 pages
...one fourth of this remainder must be the Bum of the given series ; but 4 is the ratio less 1. Hence, Rule. Multiply the last term, by the ratio, from 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.... | |
 | William Seneca Sutton - 1896 - 342 pages
...PROCESS. 3 x 243 = 729 ; 729 - 3 = 726 ; J of 726 = 363. To find the sum of a geometrical series : 357. Multiply the last term by the ratio. From the product subtract the first term. Divide the remainder by the ratio less 1. 2. The first term is 5 ; the multiplier is 6 ; the last term... | |
 | Joseph Ray - Algebra - 1848 - 260 pages
...s= - j— = — r . *•— 1 r— 1 Hence, the BULK, FOR FINDING THE SCSI 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, (i, 18, 54, Jto. The hwt term... | |
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