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. Elements of algebra - Side 190av Bourdon (Louis Pierre Marie, M.) - 1838 - 355 siderUten tilgangsbegrensning - Om denne boken
| Charles Hutton - 1815
...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 - 264 sider
...= 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 - 270 sider
...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 - 288 sider
...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 - 252 sider
...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 - 311 sider
...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 - 358 sider
...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 - 258 sider
...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 - 1843 - 12 sider
...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 - 340 sider
...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... | |
| |