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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. "
Practical and Mental Arithmetic ... - Σελίδα 263
των Roswell Chamberlain Smith - 1839
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Bradbury's Eaton's Practical Arithmetic, Combining Oral and Written Exercises

William Frothingham Bradbury - 1879
...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....

New Elementary Algebra: Designed for Common and High Schools and Academies

Shelton Palmer Sanford - 1879 - 332 σελίδες
...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...

New Elementary Algebra: Designed for the Use of High Schools and ..., Βιβλίο 1

Benjamin Greenleaf - 1879 - 336 σελίδες
...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...

New Practical Algebra: Adapted to the Improved Methods of Instruction in ...

James Bates Thomson - 1880 - 312 σελίδες
...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...

Bradbury's Eaton's Practical Arithmetic, Combining Oral and Written ...

William Frothingham Bradbury - 1881
...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....

New Practical Algebra

James Bates Thomson - 1884
...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....

Practical Arithmetic, by Induction and Analysis, Βιβλίο 3

Joseph Ray - 1885 - 336 σελίδες
...Rule for Case II. — Multiply the greatest term by the ratio ; from theproduct subtract the least term, and divide the remainder by the ratio less 1 ; the quotient will be the sum of the series. NOTE. — When a series is decreasing, and the number of terms infinite, the last term is naught. In...

Numbers Applied, a Complete Arithmetic, Μέρος 1

Andrew Jackson Rickoff - 1886
...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...

Numbers Applied

Andrew Jackson Rickoff - 1888 - 235 σελίδες
...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...

The Academic Algebra

William Frothingham Bradbury, Grenville C. Emery - 1889 - 414 σελίδες
...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...




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