 | Joseph Ray - 1857 - 358 σελίδες
...Role for Case II. — Multiply the greatest term by the ratio ; from the product 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... | |
 | 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... | |
 | Thomas Tucker Smiley - 1868 - 238 σελίδες
...that product will be the last term. 3. Multiply the last term by the ratio; from the product fcubtract the first term, and divide the remainder by the ratio, less 1, for the sum of the series. Question». What is Geometrical Progression? What № the ratio ? By what... | |
 | 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...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 is... | |
 | 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... | |
 | 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 of... | |
 | 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...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 of... | |
 | 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... | |
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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. 
