And (141) the compounded ratio is 12 X 4 X 8 X 16 tion in its lowest terms is, denoting the ratio of 1 to 32. which frac 146. All the terms of a geometrical progression may be expressed by means of the common ratio and one of the extremes. Thus, the series 3, 6, 12. 24, 48, &c. where the common ratio is 2, and first term 3, will be 3, 3 x 2, 3 X 2 × 2, 3 × 2 × 2 × 2, 3 × 2 × 2 × 2 × 2, &c. or 3, 3 x 2, 3 × 22, 3 × 23, 3 x 2+, &c. (111) 147. Therefore in any ascending progression, if the first term be multiplied by the ratio raised to the power whose index is the number of terms less by 1, the product will be the last term. For example, suppose the first term is, the common ratio 2, and the number of terins 18; what is the last term? The number of terms less by 1 is 9: last term. And 29 512, which multiplied by (the first term) gives 128 the 148. But in a descending progression (where the terms result from division) the first term divided by the said power of the ratio gives the last term. Thus, suppose the first term is 128, the common ratio 2, and the number of terms 10; what is the last term? 29512; and 128 divided by 512 gives or (in its lowest terms) the last term. 149. Hence, if one extreme be divided by the other, the quotient will be that power of the ratio whose index is the number of terms less by 1; and consequently its root will be the ratio. For example, if 7 be the first term, 189 the last, and 4 the number of terms; what is the ratio? 18227 the 3d. power of the ratio (the number of terms being 4}, whose cube root is 3 the ratio required. Therefore the 4 terms are 7, 7 x 3, 7 x 32, 189. 150. In like manner we find a proposed number of geometrical mean proportionals between two given numbers. For example, let it be required to find 3 geometrical means between 6 and 1536. 1536 256 the 4th. power of the ratio (the nuinber of terms being 5). The square root of 256 is 16 whose square root is 4, the 4th root of 256 or the required ratio. And the three means will be 6 × 4, 6 × 42, 6 × 43; 151. When only one mean proportional between two given numbers is required, the square root of their product will be the answer. For example, to find a mean proportional between 8 and 18. 8 × 18 = 144 whose square root is 12 the answer. For 8 12 12 18. And 18 is called a third proportional to 8 and 12. 152. To find the sum of all the terms in a given progression; suppose 2, 6, 18, 54, 162; where the common ratio is 3. 2, 6, 18, 54, 162 3 6, 18, 54, 162, 486 the series multiplied by the ratio 3. 2. 6, 18, 54, 162. This remainder is equal to twice the sum of the series, because it is the difference between the series and three times the series. Therefore if 486 less by 2, be divided by 2 (viz. the ratio less by 1) the quotient will be the sum of the series. But 486 less by 2 is the difference between the first term, and the product of the last by the ratio: hence the following Rule. Multiply the last term by the ratio, and take the first term from the product, then divide the difference by the ratio lessened by 1, and the quotient is the sum of the progression. In a descending progression take the first term for the last, and vice versa. Ex. 2. Required the sum of the series 65536, 16384, 4096, &c. continued to 12 terms? The ratio or divisor is 4; and 4" 4194304: And 65536 divided by 4194304 gives 13 or (in its lowest terms) the 12th, or last term of the series, which being made the first term, and 65536 the last, the work will stand as below. 3. An officer with a detachment of 60 men having taken a very strong fort by surprize, desired as a reward for himself and the party, 1 musket bullet for the first man, 2 for the second, 4 for the third, 8 for the fourth, and so on, doubling to 60 times (the number of men). Now suppose each bullet to be an ounce, and the lead at 5 shillings the hundred weight; what would be the value of his request? Here the first term is 1, the ratio 2, and the number of terms 60; therefore 259, or 2 raised to the 59th. power will be the last term of the series. The 6th. power of 2 is 64, which cubed is 262144 the 18th. power (111) and that cubed gives 18014398509481984 the 54th, power, which multiplied by 32 (the 5th. power of 2) is 576460752303423488 the 59th. power or last term of the series; this multiplied by 2 the ratio, and 1 (the first term) subtracted from the product, gives 1152921504606846975 the sum of the series, or number of bullets, or ounces (because the ratio lessened by 1 is 1), equal to 643371375338642, hundred weight, which at 5 shillings the hundred, amounts to £160842843834660 the answer. |