... that is, Any term of a geometric series is equal to the product of the first term, by the ratio raised to a power, whose exponent is one less than the number of terms. EXAMPLES. 1. New Elementary Algebra - Page 305by Benjamin Greenleaf - 1866Full view - About this book
| Charles Davies - Algebra - 1835 - 378 pages
...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. For example, the 8lh term of the progression 2 : 6 :. 18 : 54 . . ., is equal to 2x37=2x2187=4374.... | |
| Algebra - 1838 - 372 pages
...it. 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. 1. Find the 5th term of the progression 2 : 4 : 8 : 16, &c, in which the first term is 2 and the common... | |
| Charles Davies - Arithmetic - 1838 - 292 pages
...first term, the common ratio, and the number of terms, to find the last term. RULE. Raise the ratio to a power whose exponent is one less than the number of terms, and then multiply the power by the first term, the product will be the last term. EXAMPLES. 1. The... | |
| Charles Davies - Algebra - 1839 - 272 pages
...precede it. Hence, to find the last term of a progression, we have the following RULE. I. Raise the ratio to a power whose exponent is one less than the number of terms. II. Multiply the power thus found by the first term: the product will be the required term. QUEST.... | |
| Charles Davies - Algebra - 1842 - 368 pages
...it. That is, the last term of a geometrical progression is equal to the f,rst term multiplied by the ratio raised to a power whose exponent is one less than the number of terms. 1. Find the 5th term of the progression 2:4:8:16, &c, in which the first term is 2 and the common ratio... | |
| Charles Davies - Algebra - 1842 - 284 pages
...precede it. Hence, to find the last term of a progression, we have the following RULE. I. Raise the ratio to a power whose exponent is one less than the number of terms. II. Multiply the power thus found by the first term: the product will be the required term. QUEST.... | |
| Charles DAVIES (LL.D.) - Arithmetic - 1843 - 348 pages
...first term, the common ratio, and the number of terms, to find the last term. RULE. Raise the ratio to a power whose exponent is one less than the number of terms, and then multiply the power by the first term, the product will be the last term. EXAMPLES. 1. The... | |
| Charles Davies - Arithmetic - 1844 - 666 pages
...first term, the common ratio, and the number of terms, to find the last term. RU1E. Raise the ratio to a power whose exponent is one less than the number of terms, and then multiply the power by the fa-st term, the product mil be the last term. EXAMPLES. 1. The first... | |
| Charles Davies - Arithmetic - 1846 - 370 pages
...given the first term, the common ratio, and the number of terms, to find the last term, Raise the ratio to a power whose exponent is one less than the number of terms, and then multiply the power by the first term : the product will be the last term. EXAMPLES. 1 . The... | |
| George Roberts Perkins - Arithmetic - 1846 - 266 pages
...of terms, to find the first term, we have this RULE. Divide the last term by a power of the ratio, whose exponent is one less than the number of terms. EXAMPLES. 1. The last term of a geometrical progression IB 1048576, the ratio is 4, and the number of terms is 11.'... | |
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