 That is, the last term of a geometrical progression is equal to the first term multiplied by the ratio raised to a power whose exponent is one less than the number of terms.  Elements of Algebra - Page 1901839 - 355 pagesFull view - About this book
 | William Guy Peck - Algebra - 1875 - 348 pages
...to ar 3 ; and so on to the n* term, which is equal to ar n ~ l ; hence, I = ar"- 1 (1) that is, any term of a geometrical progression is equal to the first term, multiplied by that power of the ratio whose exponent is equal to the number of preceding terms. EXAMPLES. 1. Find... | |
 | William Guy Peck - Algebra - 1875 - 348 pages
...to ar3; and so on to the пл term, which is equal to ar""1; hence, I = ar"-1 ..... (1) that is, any term of a geometrical progression is equal to the first term, multiplied bу that power of the ratio whose exponent is equal to the number of preceding terms. EXAMPLES. 1.... | |
 | William James Milne - Arithmetic - 1877 - 402 pages
...X 2-, the fourth 3 X 22 X 2 or 3 X 23 and " ' ' the fifth 3X23X2 or 3 X 2<, that is, the fflh term is equal to the first term multiplied by the ratio raised to the fourth power. RULE. — Any term of a geometrical progression is equal to the first term, multiplied... | |
 | William James Milne - Arithmetic - 1878 - 406 pages
...term is equal to the first term multiplied by the ratio raised to the fourth power. RULE.- — Any term of a geometrical progression is equal to the first term, multiplied by thn ratio raised to a power one less than the number of the term. 3. The first term of a geometrical... | |
 | George E. Seymour - Arithmetic - 1880 - 332 pages
...the fifth power of the ratio. GENERAL LAW OF THE SERIES. 539. Any term in a geometrical series equals the first term, multiplied by the ratio raised to a power whose index is one less than the number of the term. Find the last term in the following series : 2. First... | |
 | Charles Davies - Algebra - 1889 - 330 pages
...ar3 ; and so on to the wth term, which is equal to ar"~l; hence, I = ar"~l ...... (1.) That is, any term of a geometrical progression is equal to the first term, multiplied by that power of the ratio whose exponent is equal to the number of preceding terms. EXAMPLES. 1. Find... | |
 | William James Milne - Algebra - 1892 - 370 pages
...the first term, r the ratio, and I the last or wth term, I. I = ar—1. That is, 356i PRINCIPLE. — The last term of a geometrical progression is equal...term, multiplied by the ratio raised to a power whose index is 1 less than the number of terms. EXAMPLES. 1. Find the 8th term of the series 2, 4, 8, etc.... | |
 | William James Milne - Arithmetic - 1892 - 440 pages
...22, the fourth 3 x 22 x 2 or 3 x 2», and the fifth 3 x 2« x 2 or 3 x 24, that is, the fifth term is equal to the first term multiplied by the ratio raised to the fourth power. RULE. — Any term of a geometrical progression is equal to the first term, multiplied... | |
 | John Tilden Prince - Arithmetic - 1894 - 246 pages
...upon the above numbers, the following rules may be developed : The last term of a geometrical series is equal to the first term multiplied by the ratio raised to a power whose degree is one less than the number of terms. The first term is equal to the last term divided by the... | |
 | William James Milne - Algebra - 1906 - 444 pages
...represents the first term, r the ratio, and I the last or nth term, I. l=arn~1. That is, 356. PRINCIPLE. — The last term of a geometrical progression is equal...the first term, multiplied by the ratio raised to a vower whose index is 1 less than the number of terms. EXAMPLES. 1. Find the 8th term of the series... | |
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