 | Roswell Chamberlain Smith - Arithmetic - 1856 - 334 pages
...series, we have the following RULE. 21. Multiply the last term by the ratio, from the product subtrac the first term, and divide the remainder by the ratio, less 1 ; the qua tient will be the sum of the series required. 22. If the extremes be 5 and 6,400, and the... | |
 | Charles Guilford Burnham - Arithmetic - 1857 - 328 pages
...find the sum of the series, we have the following RULES. I. Multiply the last term by the ratio, and from the product subtract the first term, and divide the remainder by the ratio less 1; the quotient will be the answer. II. Divide the difference between the two extremes by the ratio less... | |
 | Charles Guilford Burnham - 1857 - 342 pages
...find the sum of the series, we have the following RULES. I. Multiply the last term by the ratio, and from the product subtract the first term, and divide the remainder by the ratio less 1; the quotient will be the answer. II. Divide the difference between the two extremes by the ratio less... | |
 | Joseph Ray - Algebra - 1857 - 408 pages
...— 1 ' — 1 This formula gives the following RULE, FOR FINDING THE SUM OF A GEOMETRICAL SERIES. — Multiply the last term by the ratio, from the product subtract the first term,and divide the remainder by the ratio less one. Ex. Find the sum of 6 terms of the progression... | |
 | Elias Loomis - Algebra - 1862 - 312 pages
...terms of a geometrical progression, we have the following RULE. Multiply the last term by the ratio, subtract the first term, and divide the remainder by the ratio less one. Examples. 1. What is the sum of nine terms of the series 1, 3, 9, 27, 81, etc. ? We have already... | |
 | Benjamin Greenleaf - 1863 - 338 pages
...value of ar" in (4), S — rl — a. r-4 '-7^1 . <•'•) Hence the RULE. die last term by 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... | |
 | Joseph Ray - Algebra - 1866 - 252 pages
...H'ow 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 =2X3' =2 X 19683=39366.... | |
 | Joseph Ray - Algebra - 1866 - 420 pages
...— a Therefore ...... 8= - =- = - =-. Hence, 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. Find the sum of 6 terms of the progression 3, 12, 48, etc. 3=4095, Ans. order that both terms... | |
 | Horatio Nelson Robinson - Algebra - 1866 - 328 pages
...— 1)8 = rL — a. Or, 8 = ^. « Hence, the following 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 PRACTICE. 1. The first term is 5, the last term 1280, and the ratio 4 ; what is the... | |
 | William Frothingham Bradbury - Algebra - 1868 - 264 pages
......... -f' + Zr(2) Subtracting (1) from (2), rS — S = I r — a lr _ a Whence, S= - — . Hence, RULE. ' Multiply the last term by the ratio, from...first term, and divide the remainder by the ratio less one. 1. Given a = 2, I= 20000, and r = 10, to find S. lr — a _ 20000 X 10 — 2 _ gf>p . S — -jr--r... | |
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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. 
