? each of its terms by the ratio, a new series will be formed, of which every term, except the last, will have its corresponding term in the first series. Thus, take the series 1, 3, 9, 27, 81, the sum of which we will represent by S, so that S=1+3+9+27+81. Multiplying each term by 3, we obtain 3S=3+9+27+81+243. The terms of the two series are identical, except the first term of the first series and the last term of the second series. If then we subtract one of these equations from the other, all the remaining terms will disappear, and we shall have 3S-S-243-1, or 243-1 S= 3-1 (212.) In order to generalize this method, let a, ar, ar3, etc., represent any geometrical series, the last term of which is 7, and let S represent the sum of all the terms; then +lr. Multiplying this equation by r, we obtain rS=ar+ar+ar3 +ar1+ Subtracting the first equation from the second, all the terms in the second members disappear, except the first term of the first series and the last term of the second series, and we obtain (213.) Hence, to find the sum of the terms of a geometrical progression, we have the following RULE. Multiply the last term by the ratio, subtract the first term, and divide the remainder by the ratio less one. Examples. 1. What is the sum of nine terms of the series 1, 3, 9, 27, 81, etc.? We have already found the ninth term of the series 2. What is the sum of eight terms of the series 1, 4, 16, 64, etc.? Ans. 21845. 3. What is the sum of 14 terms of the series 1, 2, 4, 8, 16, etc.? Ans. 16383. 4. The extremes of a geometrical progression are 3 and 12288, and the common ratio 2. What is the sum of the series? Ans. 24573. 5. The extremes of a geometrical progression are 4 and 78732, and the common ratio 3. What is the sum of the series? Ans. 118096. 6. What debt may be discharged in 12 months by QUEST.-Give the rule for the sum of the terms of a geometrical progression. Explain the reason of the rule. paying one dollar the first month, two dollars the second month, four dollars the third, and so on, each succeeding payment being double the last; and what will be the last payment? Ans. the debt is $4095, the last payment $2048. (214.) When the progression is a decreasing one, and r consequently represents a fraction, the expression for the sum of the series is written a-Ir in order that both terms of the fraction may be positive. Ex. 1. What is the sum of 9 terms of the progression 1536, 768, 384, etc.? 1536-3 Here 7=1536×(1)=6. Hence S= =3066. Ex. 2. What is the sum of 11 terms of the progres5120, 2560, 1280, etc.? sion Ans. 10235. Ex. 3. What is the sum of 12 terms of the progression 8192, 4096, 2048, etc.? Ans. 16380. Ex. 4. What is the sum of-7 terms of the progression 15625, 3125, 625, etc.? Ans. 19531. Ex. 5. What is the sum of 8 terms of the progression 32768, 8192, 2048, etc.? Ans. 43690. Ex. 6. What is the sum of 9 terms of the progression 19683, 6561, 2187, etc.? Ans. 29523. QUEST.-What is the sum of the terms of a decreasing progression? PROBLEM VI. (215.) To find a mean proportional between two numbers. According to Art. 187, if b is a mean proportional between a and c, we shall have that is, to find a mean proportional between two numbers, multiply the two numbers together, and extract the square root of their product. Ex. 1. What is the geometrical mean between the numbers 4 and 9? Ans. √4x9=6. Ex. 2. What is the geometrical mean between the numbers 4 and 25? Ans. 10. Ex. 3. What is the geometrical mean between the numbers 9 and 16? Ans. 12. Ex. 4. What is the geometrical mean between the numbers 4 and 49? Ans. 14. Ex. 5. What is the geometrical mean between the numbers 9 and 25? Ans. 15. Ex. 6. What is the geometrical mean between the numbers 4 and 81? Ans. 18. QUEST.-HOW may we find a mean proportional between two num. bers? Ex. 7. What is the geometrical mean between the numbers 16 and 25? Ans. 20. (216.) Of decreasing progressions having an infinite number of terms. In a decreasing progression, the expression for the sum of the series is For substitute its value ara-1, and we obtain In a decreasing progression, sincer is less than unity, r is less than unity; and the larger the number n, the smaller will be the quantity r". If, therefore, we take a very large number of terms of the series, the arn quantity, and consequently the term 1-,, will be very small; and if we take n greater than any assign arn able number, then will be less than any assign 1-r able number, and may be neglected; that is, when the number of terms is infinite, we have (217.) Hence the sum of an infinite series decreasing in geometrical progression is found by the fol lowing |