There are twenty different cases in Arithmetical Progression, all of which are exhibited in the following TABLE. 12 Formulæ. l=a+(n−1)d. l=d±2dS+(a−d)3. n + S=n[2a+(n−1)d]. S= + s=(1+a)n. 2 2d S=±n(21—(n−1)d). SECTION XXI. GEOMETRICAL PROGRESSION, OR PROGRESSION BY ART. 269. When there are three or more numbers, such that the same quotient is obtained by dividing the second by the first, and the third by the second, and the fourth by the third, &c.; or, such that they increase or decrease by a constant multiplier, they are said to be in Geometrical Progression, and are called a Geometrical Series. Thus, (1) (2) (3) (4) (5) (6) (1.) 2, 6, 18, 54, 162, 486 = 728, sum of the series. (2.) 486, 162, 54, 18, 6, 2= 728, sum of the series. The first is called an ascending series, and the second a descending series. In the first the quotient or multiplier is 3, and it is called the ratio. In the second the ratio is . 270. The first and last terms of a series are called the extremes, and the others are the means. 271. It will readily be perceived, in either of the above series, that the product of the extremes is equal to the product of any two of the means equally distant from the extremes. Thus, 2×486=6×162=18×54=972. 272. If there are only three terms, the product of the extremes is equal to the square of the second term. 273. It is evident, by examining either the above series, that any term may be obtained by multiplying the first term by the ratio as many times, wanting' one, as there are terms required. If, therefore, the 1st term is 2, and the ratio 3, and we wish to obtain the 6th term, we have only to multiply the 1st term, 2, by the ratio 3, five times. Thus, 2×3×3×3×3×3=486, the 6th term. The above may be generalized in the following manner : Let a first term of a series, L= the last term. Then a, ar, ar2, ar3, art, ar, &c., may represent any geometrical series; and, if r, the ratio, is considered as more than a unit, the series is ascending; but, if r is less than a unit, the series is descending. The exponent of r in the second term is 1, in the third term 2, in the fourth term 3, in the fifth term 4, and so on; therefore, the exponent of r in the last term will always be one less than the number of terms. The exponent of the nth term in the above series would therefore be ar"-1. 274. If, therefore, in any series the number of terms be denoted by n, and the last term by L, the following will be the formula for finding the last term: (1.) And L=ar”-1. L="-1, when the first term is a unit. In the above equation we have four quantities, a, L, r, and n; and, if any three of them be given, the others may be obtained as follows: To find a, the first term, we divide both terms of the above equation by -1, and transpose the terms; and we have To obtain r, the ratio, we divide the terms of the 1st equation by a, extract the (n-1)th root, and transpose the terms; and we have To find n, we shall show when we come to treat of exponential quantities. 1. If the first term is 7, the ratio 3, and the number of terms 5, required the last term. 2. If the first term is 1, the ratio 5, and the number of terms 5, what is the last term? 3. If the last term is 405, the ratio 3, and the number of terms 5, what is the first term? 4. If the last term is 8, ratio 5, and the number of terms 4, what is the first term? 5. If the first term is 5, the last term 1215, and the number of terms 6, what is the ratio? 27 320' 6. If the first term is, the last term and the number of 7. If the first term is, the last term 64, and the number of terms 6, required the ratio. Ans. 4. 8. If the last term is 135, the number of terms 4, the ratio 3, what is the first term? Ans. 5. 275. To find any number of geometrical means between any two given numbers. n-1 In the 3d formula, we found r= a If we let m represent the number of means, then m+2=n; for the number of terms is always two more than the number 276. Having, therefore, the extremes given to find any number of means, we divide the greater extreme or number by the less extreme, and extract that root of the quotient denoted by the number of means plus 1. This root is the ratio; and, having the ratio, the means are readily obtained. EXAMPLES. 9. Find two geometrical means between 6 and 162. 162÷6=27:27=3, the ratio; 6×3=18, the first mean; 18x3=54, the second mean. 10. What is the geometrical mean between 18 and 882? 882÷18=49:49=7, the ratio; 18x7=126, the geometrical mean. 11. Required the five geometrical means between 1 and 64. Ans. 2, 4, 8, 16, 32. 12. A has a piece of land, which is 18 rods wide, and 288 rods long. Required the side of a square piece that shall contain an equal number of square rods. Ans. 72 rods. 277. To find the sum of all the terms of a geometrical series. Let the following be the series: By examining this series, we find the first term 2, the ratio 3, and the last term 162. If we multiply each term in the series by the ratio 3, we obtain It is evident that the sum of this last series is three times the |