 | Charles Davies - 1835 - 378 páginas
...values of all those which precede it. That is, the last term of a geometrical progression is equal to the first term multiplied ' by the ratio raised to...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.... | |
 | 1838 - 372 páginas
...find all the terms which precede it. That is, the last term of a geometrical progression is equal to the first term multiplied by the ratio raised to a...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 - 1838 - 292 páginas
...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 - 1839 - 272 páginas
...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 - 1842 - 368 páginas
...the terms which precede 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 - 1842 - 284 páginas
...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.) - 1843 - 348 páginas
...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 - 1844 - 666 páginas
...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 - 1845 - 382 páginas
...to find all the terms which precede it. That is, Any term of a geometrical progression is equal to the first term multiplied by the ratio raised to a power whose exponent denotes the number of preceding terms. EXAMPLES. 1. Find the 5th term of the progression 2 : 4 : 8... | |
 | Charles Davies - 1846 - 370 páginas
...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... | |
| |
... 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. 
