 | Lorenzo Fairbanks - 1875 - 472 σελίδες
...to a power whose index is equal to the number of terms, and multiply this power by the first term ; from the product subtract the first term, and divide the remainder by the ratio less 1. NOTE. — This rule is employed in computing the values of annuities. EXAMPLES. 1. The first term... | |
 | James E. Ryan - 1877 - 212 σελίδες
...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 the product...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 many... | |
 | Albert Newton Raub - 1877 - 348 σελίδες
...of a geometrical series, as follows : RULE. Multiply the last term by the ratio, a.nd from this take the first term, and divide the remainder by the ratio, less one. The following is the formula : S. = - . r—\ 2. A clerk whose first year's salary is $75 has his salary... | |
 | William Frothingham Bradbury - 1877 - 280 σελίδες
...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
...in the given series. Substituting ¡r for ar", we have FOKMULA II. s = — — • r— i RULE. — Multiply the last term by the ratio, from the product...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 a = 2,... | |
 | 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...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. 4372. 107.... | |
 | 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...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. • 107.... | |
 | Benjamin Greenleaf - 1879 - 350 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... | |
 | Shelton Palmer Sanford - 1879 - 348 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... | |
 | William Frothingham Bradbury - 1881 - 404 σελίδες
...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. 107.... | |
| |
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. 
