 | 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.... | |
 | Daniel W. Fish - 1874 - 300 σελίδες
...hence, ^,or 170, i|4 = 170 = * is the sum. 2. The extremes are 1 and {\, and the ratio is K ; what is the sum of all the terms ? RULE. — Multiply the last term by the ratio, and divide the difference between the product and the first term by the difference between 1 and the... | |
 | William Frothingham Bradbury - 1875 - 280 σελίδες
...rS=ar+ar,-\- ar* + _|-i_|-ir (2) Subtracting (1) from (2), r S — A = Ir — a Whence, S = . Hence, RULE. Multiply the last term by the ratio, from the product subtract tlw first term, and divide the remainder by the ratio less one. 1. Given a = 2, I = 20000, and r= 10,... | |
 | James E. Ryan - 1877 - 212 σελίδες
...equals be divided by (r — 1), s will stand 7j« ft equal to (lr— a)-f-(r— l),or, as above, s= .-. RULE. — Multiply the last term by the ratio, from...first term, and divide the remainder by the ratio less 1 . EXERCISE CXLIV. 1,2. How many ancestors hast thou, reckoning 20 generations back ; and how... | |
 | William Frothingham Bradbury - 1877 - 280 σελίδες
...г, rS=ar+ari + ar, + -\-l-\-lr (¥) Subtracting (1) from (2), r S — S=lr — a Whence, S= . Hence, RULE. Multiply the last term by the ratio, from the product subtract the ßrst term, and divide the remainder by tlie ratio less one. 1. Given a = 2, I = 20000, and r — 10,... | |
 | James Bates Thomson - 1878 - 322 σελίδες
...last term in the given series. Substituting ¡r for ar", we have FOKMULA II. s = — — • r— i RULE. — Multiply the last term by the ratio, from...first term, and divide the remainder by the ratio less one. fW For the method of finding the sum of an infinite descending series, see Art. 435. 1. Given... | |
 | William Frothingham Bradbury - 1882 - 416 σελίδες
...one fourth of this remainder must be the sum 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 Frothingham Bradbury - 1879 - 392 σελίδες
...fourth of this remainder must be the sum of the given series ; but 4 is the ratio less 1. Hence, Bule. Multiply the last term by the ratio, from the product...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 1 Ans.... | |
 | William Frothingham Bradbury - 1879 - 446 σελίδες
...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.... | |
 | James Bates Thomson - 1880 - 324 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... | |
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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. 
