| Nicolas Pike - Algebra - 1808 - 470 pages
...the other extreme. RULE — Multiply, or divide, (as the case may require) the given extreme by such power of the ratio, whose exponent* is equal to the number of terms lefs J, and the produdl or quotient, will be the other extreme. EXAMPLES. ever, as 2. 6. 18. 54. 162.... | |
| Benjamin Peirce - Algebra - 1837 - 300 pages
...— S = arn — a, or (r — 1) S= ar» — a = a(rn — 1); whence a r» — aa (rn — 1) r — 1 " r — 1 Hence, to find the sum, multiply the first...and the ratio. Examples in Geometrical Progression. 186. Corollary. The two equations • (r—l)S = a(r»— 1) give the means of determining either two... | |
| Benjamin Peirce - Algebra - 1837 - 300 pages
...— a, or (r — I) S= ar*— a = a (r» — 1); - .. , -<•: .. whence arn — aa r» — 1 8 — Hence, to find the sum, multiply the first term by...and the ratio. Examples in Geometrical Progression. 186. Corollary. The two equations I = ar»-1 give the means of determining either two of the quantities... | |
| Benjamin Greenleaf - Arithmetic - 1839 - 356 pages
...terms, to find the other extreme. RULE. If the series be ascending, multiply the given extreme by such power of the ratio, whose exponent is equal to the number of * This rule cannot well be demonstrated except by Algebra SECT. L vI.] GEOMETRICAL SERIES. 213 terms... | |
| George Roberts Perkins - Arithmetic - 1841 - 274 pages
...the terms, repeated as many times as there are units in the number of terms, less one, equal to the power of the ratio whose exponent is equal to the number of terms, diminished by one and multiplied by the first term. Hence, when we have given the first term, the ratio,... | |
| Benjamin Greenleaf - Arithmetic - 1841 - 334 pages
...terms, to find the other extreme. RULE. If the series be ascending, multiply the given extreine by such power of the ratio, whose exponent is equal to the number of * Thii rule cannot well be demonstrated except by Algebra terms lesp 1 ; and the product will be the... | |
| George Roberts Perkins - Arithmetic - 1846 - 266 pages
...the terms, repeated as many times as there are units in the number of terms, less one, equal to the power of the ratio whose exponent is equal to the number of terms, diminished by one, and multiplied by the first term. Hence, when we have given the first term, the... | |
| George Roberts Perkins - Arithmetic - 1849 - 344 pages
...1398101 pecks. Case III. Since by Case I. the last term is equal to the first term multiplied into a power of the ratio whose exponent is equal to the number of terms, less one, it follows that the first term is equal to the last term divided by the power of the ratio... | |
| George Roberts Perkins - Arithmetic - 1850 - 356 pages
...the terms, repeated as many times as there are units in the number of terms, less one, equal to the power of the ratio, whose exponent is equal to the number of terms diminished by one, and multiplied by the first term. Hence, when we have given the first term, the... | |
| Benjamin Peirce - Algebra - 1851 - 294 pages
...n, or (r — 1) 8 = ar n — a = a (r n — 1) ; whence o = - = —. a r" — an(r n — 1) r — 1 r — 1 Hence, to find the sum, multiply the first...Geometrical Progression. 260. Corollary. The two equations (r — l}S = a(r" — 1) give the means of determining either two of the quantities a, I, r, n, and... | |
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