 | George Roberts Perkins - 1841 - 274 Seiten
...multiplied by the third power of the ratio ; and so on, for the succeeding terms. Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this . RULE. Multiply the first term by the power of the ratio, whose exponent it... | |
 | Benjamin Greenleaf - 1843 - 340 Seiten
...cent. for 30 years ? Ans. $574.34.91172913250116264106332310802645846357252196069357387776. PROBLEM IL Given the first term, the ratio, and the number of terms, to find the sura of the series. RULE. Find the last term, as before, multiply it by the ratio, and from the product... | |
 | John Darby (teacher of mathematics.) - 1843 - 236 Seiten
...16, 64, and the equidistant terms, are in geometrical progression, whose ratio is 4. PROBLEM I. — Given the first term, the ratio, and the number of terms to find the last or any other assigned term. RULE — Find such a power of the ratio as is denoted by the number... | |
 | mrs. Henry Ayres - 1843 - 470 Seiten
...it, but that belongs more particularly to Geometrical Progression. CASE I. — Given the extremes of the number of terms, to find the sum of all the terms. Rule. — Add the extremes together, multiply by the number of terms, and divide by 2. Examples. \. The extremes... | |
 | Charles Davies - 1844 - 110 Seiten
...We have /=lX3"=lx 19683=19683 cents. , Iq-a 19683X3-1 y — 1 =29524 cents. In this example we have the first term, the ratio, and the number of terms, to find the last term and the sum of the series. J=4X 815=4 X 35184372088832 = 140737488355328. For the sum of... | |
 | Charles Waterhouse - 1844 - 232 Seiten
...easily understood. In the first case and its rule in the Table, read — 1. Given, the first term, ratio, and the number of terms, to find the sum of all the terms ; or, the last term. RULE 1. — The ratio less 1, raised to the power denoted by the number of terms,... | |
 | George Roberts Perkins - 1846 - 266 Seiten
...multiplied by the third power of the ratio ; and so on, for the succeeding terms. Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this RULE. Multiply the first term by the power of the ratio, whose exponent is... | |
 | mrs. Henry Ayres - 1846 - 400 Seiten
...it, but that belongs more particularly to Geometrical Progression. CASE I. — Given the extremes of the number of terms, to find the sum of all the terms. Rule. — Add the extremes together, multiply by the number of terms, and divide by 2. Examples. 1. The extremes... | |
 | William Vogdes - 1847 - 324 Seiten
...the oldest son's portion ? § 173« CASE 3. The first term, the number of terms, and the ratio given, to find the sum of all the terms. RULE. Find the last term, as before, then subtract the first from it, and divide the remainder by the ratio less one, to the... | |
 | George Roberts Perkins - 1849 - 356 Seiten
...multiplied by the third power of the ratio ; and so on, for the succeeding terms. Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this RULE. Multiply the first term by the power of the ratio whose exponent is one... | |
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Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this RULE. 
