 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.  Stoddard's Practical Arithmetic - Page 233by John Fair Stoddard - 1852 - 299 pagesFull view - About this book
 | Nathan Daboll - Arithmetic - 1843 - 254 pages
...first term, the last term, (or the extremes,) and the ratio given, to find the sum of the series. BULB. Multiply the last term by the ratio ; from the product...first term, and divide the remainder by the ratio, less 1, and the quotient will be the sum of all the terms. EXAMPLES. 1 . A man bought 6 yards of cloth,... | |
 | Roswell Chamberlain Smith - Arithmetic - 1843 - 310 pages
.....- .- >• . • :•••••": •> RULE. _ ........ ' ., • f, g 1'iir1:' ' • • • » J Multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio, less 1 ; the quotient will be the sum of the series required. . . , 9. If the extreme! be 5 and 6400,... | |
 | James Bates Thomson - Algebra - 1844 - 272 pages
...term in the given series. 373. To find the sum of a geometrical series. Multiply the last term into the ratio, from the product subtract the first term, and divide the remainder by the ratio less one. Obser. From the above formula, in connexion with the one in Art. 368, there may be the same... | |
 | Arithmetic - 1845 - 196 pages
...terms given, which, being multiplied by the first term, will give the last term, or greater extreme. 2. Multiply the last term by the ratio, from the product...subtract the first term, and divide the remainder by ratio less one for the sum of the series. EXAMPLES. 1. A thresher wrought 20 days, and received for... | |
 | Joseph Ray - Algebra - 1848 - 250 pages
...Therefore, *= =— = =•. r—1 r—l Hence, the 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. EXAMPLES. 1. Find the sum of 10 terms of the progression 2, 6, 18, 54, &,1:. The last term... | |
 | Jeremiah Day, James Bates Thomson - Algebra - 1848 - 264 pages
...term in the given series. 373. To find the sum of a geometrical series. Multiply the last term into -the ratio, from the product subtract the first term, and divide the remainder by the ratio less one. Obaer. From the above formula, in connexion with the one. iu Art. 368, there may be the same... | |
 | Joseph Ray - Algebra - 1848 - 252 pages
...— — ^— = - =•. r — 1 i — l Hence, the RULE, FOR FINDING THE SUM OF A GEOMETRICAL SERIES. Multiply the last term by the ratio, from the product subtract the frst term, and divide the remainder by the ratio less one. EXAMPLES. 1. Find the sum of 10 terms of... | |
 | Nathan Daboll, David Austin Daboll - Arithmetic - 1849 - 260 pages
...last term, (or the extremes,) and the ratio given, to find the sum of the series. RULE. Multiply tke last term by the ratio ; from the product subtract...first term, and divide the remainder by the ratio, less 1, and the quotient will be the sum of all the terms. EXAMPLES. 1 . A man bought 6 yards of cloth,... | |
 | Roswell Chamberlain Smith - Arithmetic - 1850 - 314 pages
...when the extremes and ratio are given, to find the sum of the series, we have the following RULE. 21. Multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio, less 1 ; the quotient will be the sum of the series required. 22. If the extremes be 5 and 6,400, and... | |
 | Benjamin Naylor - 1850 - 334 pages
...(2) (the first term of the first.) Hence the RULE. product will be the last or greater extreme 2- — multiply the last term by the ratio, from the product...first term, and divide the remainder by the ratio less one for the sum of the series, or raise the ratio to a power equal to the number of terms ; subtract... | |
| |
