 | Horatio Nelson Robinson - 1859 - 368 σελίδες
...of a geometrical series ; RULE. Multiply the last term by the ratio, and from the vroduct iub/ract 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 94erms of the series,... | |
 | Elias Loomis - 1862 - 312 σελίδες
...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 found... | |
 | Benjamin Greenleaf - 1863 - 338 σελίδες
...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 - 1866 - 420 σελίδες
...— a Therefore ...... 8= - =- = - =-. Hence, Rule for finding the Sum of a Geometrical Series. — Multiply the last term by the ratio, from the product...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 of the... | |
 | Joseph Ray - 1866 - 252 σελίδες
...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 the product...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.... | |
 | Horatio Nelson Robinson - 1866 - 328 σελίδες
...— 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 sum... | |
 | William Frothingham Bradbury - 1868 - 264 σελίδες
...-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 the product...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... | |
 | William Frothingham Bradbury - 1868 - 270 σελίδες
...-|-Z-|-Zr (3) Subtracting (1) from (2), r S — S = / r — a Whence, S = — -p. Hence, T ^~~~ 1 RULE. Multiply the last term by the ratio, from the product subtract the first term, and divide tlie remainder by the ratio less one. 1. Given a = 2, 1= 20000, and r = 10, to find S. *=£=£ = soooo^io... | |
 | John Homer French - 1869 - 350 σελίδες
...quotient is 1. The number of divisions will be the number of terms. V. To find the sum of the series. Multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio less 1. fit OELEMS. 6. The first term of an ascending geometrical series is 7, and the ratio is 3. What... | |
 | Horatio Nelson Robinson - 1874 - 340 σελίδες
...1)S = rL — a. Or, B = T=r (*) 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 sum... | |
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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. 
