| Benjamin Peirce - 1837 - 300 σελίδες
...or nth term is / = or»-1; that is, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms. 185. Problem. To -find the sum of a geometrical progression, of which the first term, the ratio, and... | |
| Benjamin Peirce - 1837 - 302 σελίδες
...is, the last or nth term is that is, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms. 185. Problem. To find th&~ sum of a geometrical progression, of which the first term, the ratio, and... | |
| George Roberts Perkins - 1841 - 274 σελίδες
...=4096, this, diminished by one, becomes 4095, which, multiplied by 2048, becomes 8386560; again, the power of the ratio, whose exponent is one less than the number of terms, is 2048, which, multiplied by the ratio, less one, is not changed; .-. 8386560 divided by 2048, gives... | |
| George Roberts Perkins - 1846 - 266 σελίδες
...number of terms, less one, it follows that the first term is equal to the last term, divided by the power of the ratio, whose exponent is one less than the number of terms. Hence, when we have given the last term, the ratio, and the number of terms, to find the first term,... | |
| George Roberts Perkins - 1849 - 346 σελίδες
...progression is 1, the ratio is 2, and the number of terms is 7. What is the last term? In this example, the power of the ratio, whose exponent is one less than the number of terms, is 26=64, which, multiplied by the first term, 1, still remains 64, for the last term. 2. The first... | |
| George Roberts Perkins - 1850 - 364 σελίδες
...progression is 1, the ratio is 2, and the number of terms is 7. What is the last term? In this example, the power of the ratio, whose exponent is one less than the number of terms, is 26 = 64, which, multiplied by the first term. 1, still remains 64, for the last term. 2. The first... | |
| George Roberts Perkins - 1850 - 356 σελίδες
...number of terms, less one, it follows .hat the first term is equal to the last term divided by (he power of the ratio whose exponent is one less than the number of terms. Hence, when we have given the last term, the ratio, and the number of terms, to find the first term,... | |
| George Roberts Perkins - 1851 - 356 σελίδες
...progression is 1, the ratio is 2, and the number of terms is 7. What is the last term? In this example, the power of the ratio, whose exponent is one less than the number of terms, is 2 e =64, which, multiplied by the first term, 1, still remains 64, for the last term. 3. A person... | |
| Benjamin Peirce - 1855 - 308 σελίδες
...that, the last or nth term is that is, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms. 259. Problem. To find the sum of a geometrical progression, of which the first term, the ratio, and... | |
| Elias Loomis - 1855 - 356 σελίδες
...That is, 'The last term of a geometrical progression is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms. (242.) To find the sum of all the terms of a geometrical progression. If we take any geometrical series,... | |
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