 | Charles Hutton - 1815 - 686 σελίδες
...uneven number of the terms. 2, 4, 8, 16, 8421 ïti ~Гб Thus, prod. Also prod 1, 16 Tïï «, 16 ra. 3. The last term of a geometrical progression, is equal to the first term multiplied, or divided, by the ratio raised to the power whose exponent is less by 1 than the number of terms in... | |
 | William Smyth - 1830 - 278 σελίδες
...= bq = aq', d = cq = aq3 from which it will be readily inferred, that a term of any von A whatever is equal to the first term multiplied by the ratio raised to a power, the exponent of which is one less than the number, which marks the place of this term. Let L designate... | |
 | Charles Davies - 1833 - 284 σελίδες
...2=24 4th term, &c. for the other terms. But 2x2=23,2x2x2=23, and 2x2x2x2=2*. Therefore, any term of a progression is equal to the first term multiplied by the ratio raised to a power" 1 less than the number of the term. CASE I. § 230. Having given the first term, the common ratio,... | |
 | Charles Davies - 1838 - 292 σελίδες
...term, &c. for the other terms. But2x2 = 22, 2x2 x 2 = 23, and 2x2x2x2=2*. Therefore, any term of the progression is equal to the • first term multiplied by the ratio raised to a power 1 less ' than the number of the term. Q. In every Geometrical Progression, how many things are considered?... | |
 | Charles Davies - 1839 - 272 σελίδες
...which has n — 1 terms before it, is expressed by aq"~l. Let / be this term ; we then have the formula by means of which we can obtain any term without being obliged to find all the terms which precede it. Hence, to find the last term of a progression, we have the following RULE. I. Raise the ratio to a... | |
 | Calvin Tracy - 1840 - 316 σελίδες
...3, 3 is evidently to be taken as multiplier, or factor, eleven times ; that is, any term of a series is equal to the first term multiplied by the ratio raised to a power one less than the number of terms. Thus, 2 x 3 x 3 x 3x3x3x3x3x3x3x3x3" = 3x2= 354294, Ans-. or twelfth... | |
 | Charles Davies - 1842 - 368 σελίδες
...n—1 terms before it, is expressed by aq"- l . Let Z be this term ; we have the formula l=aq*- i , by means of which we can obtain any term without being...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... | |
 | Charles Davies - 1842 - 284 σελίδες
...n— 1 terms before it, is expressed by af~l. Let l be this term ; we then have the formula l=aq-\ by means of which we can obtain any term without being obliged to find all the terms which precede it. Hence, to find the last term of a progression, we have the following RULE. I. Raise the ratio to a... | |
 | Charles DAVIES (LL.D.) - 1843 - 348 σελίδες
...4th term, &c. for the other terms. But 2x2=22, 2x2x2=23, and 2x2x2x2=2*. Therefore, any term of the progression is equal to the first term multiplied by the ratio raised to a power 1 less than the number of the term. Q. In every Geometrical Progression, how many things are considered... | |
 | Charles Davies - 1844 - 666 σελίδες
...4th term, &c. for the other terms. But 2x2=22, 2x2x2=23, and 2x2x2x2=24. Therefore, any term of the progression is equal to the first term multiplied by the ratio raised to a power 1 less than the number of the term. Q. In every Geometrical Progression, how many things are considered... | |
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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. 
