stant multiplier, by which a term is derived from the preceding one, may be found by dividing any term by that which precedes it. This multiplier is called the common ratio. Thus 1, 3, 9, 27, 81, &c. is a geometrical progression, formed by the multiplication of each term by the multiplier or ratio 3. In the progression,,,, &c. the ratio is. Also, in x, -x2y, x3y3, —xy3, &c. the ratio is -xy. 98. If a be the first term, r the ratio, n the number of terms, a geometrical progression may be represented generally by In which the index of r in each term, is 1 less than the number which denotes the place of that term in the series; and the series will be an increasing or deceasing one, according as r is greater or less than unity. 1 99. If 7 be the last term, then l=ar”-1 (Art. 98). From this equation we may find any one of the quantities l, a, r, n, if the three others be known. Ex. If the first and last terms of a geometrical progression be 5 and 135, find two intermediate terms or geometric means between them. In the question a=5, l=135, n=4 .. the 2 geometric means are 15 and 45. 100. The differences of quantities in geometrical progression, are also in geometrical progres sion. If a, ar, ar2, ar3, ar1, &c. - then a-ar, ar—ar2, ar2 — ar3, ar3—ar1, &c. or, a-ar, (a-ar)r, (a—ar)r2, (a—ar)r3, &c. which last series is a geometrical progression, whose first is a — ar, and common ratio r. 101. To find the sum of a Geometrical Series. 1st. If r be greater than unity, take the upper equation from the lower one. 102. 2ndly. If r be less than unity, take the lower equation from the upper one. then a-ar"=S- rS Ex. 1. Find the sum of 8 terms of the series Ex. 2. Find the sum of 6 terms of the series . If the number of terms (n) be great, the operation is extremely tedious without the assistance of logarithms. 103. Since l=ar^-1 (Art. 99.) if any 3 of the 4 quantities, S, a, r, l, be given, the 4th can be found. a-arn 104. Since S= when r is a proper frac 1 -r tion (Art. 102); and as any such fraction when multiplied continually into itself becomes extremely small; for this reason ar" is of no assignable value with respect to the other quantity a in the numerator, when the number of terms n is indefinitely increased, therefore ar" may be rejected, Ex. 3. Find the value of the circulating deci mal 333, &c. This decimal may be represented under the form of 3 3 3 + + + &c., which is a geometrical 10 100 1000 The value of a circulating decimal may be also found by putting it S, and multiplying the equation by 10, or such multiple of 10, that the equation being taken from the product, the circulating figures of the decimal may be taken away; and by this means the exact value of S will be found. Let the last example be taken. Ex. 4. Find the value of the circulating deci mal 343434, &c. Let S 3434, &c. .. 100S=34·3434, &c. by subtracting, 99S=34 |