OPERATION. paying 6 cents the first week, 18 cents the second week, and thus to increase every week by 12 cents, till the last payment should be $ 6.18. What is the debt ? Ans. $ 162.24. 561. To find one of the extremes, when the other extreme and the number and sum of the terms are given. Ex. 1. If 3 be the first term of a series, 9 the number of terms, and 99 the sum of the series, what is the last term ? It has been shown (Art. 559) that, if the 99 x 2 -3=19, sum of the extremes be multiplied by the 9 number of terms, the product will be twice [Ans. the sum of the series; therefore, if twice the sum of the series be divided by the number of terms, the quotient will be the sum of the extremes. If from this we subtract the given extreme, the remainder must be the other extreme. RULE. — Divide twice the sum of the series by the number of terms ; from the quotient take the given term, and the remainder will be the term required. EXAMPLES 2. The sum of a series of ten thousand even numbers is 100010000, and the last term of the series is 20000. Required the first term. Ans. 2. 3. A merchant, being indebted to 22 creditors $528, ordered his clerk to pay the first $3, and the rest sums increasing in arithmetical progression. What is the difference of the payments, and the last payment ? Ans. Difference 2 ; last payment $ 45. To find any number of arithmetical means, the extremes and the number of terms being given. Ex. 1. If the first term of an arithmetical series is 1 the last term 99, and the number of terms 8, what are the second and seventh terms of the series? Ans. The second term, 15 ; the seventh, 85. We find the common difference, 14, as in Art. 557, the first term, 1, plus the common 99 - 1 14; difference, 14, gives 15 for 8 1 the second term, and the last term, 99, minus the common 1 + 14 15; 99 14 85. difference, 14, gives the sev enth term. 562. OPERATION. RULE. — Find the common difference, which, added to the less extreme, or subtracted from the greater, will give one mean. From that mean derive others in the same vay, till those required are found. EXAMPLES. 2. The extremes of a series are 4 and 49, and the number of terms 6. Required the middle two terms. Ans. 22 and 31. 3. Insert five arithmetical means between 20 and 30. Ans. 213, 23, 25, 26, and 281. GEOMETRICAL PROGRESSION. 563. Geometrical Progression, or progression by quotients, is a series of numbers that increase or decrease by a constant multiplier or divisor, called the common ratio. The series is an ascending one, when each term after the first increases by a constant ratio ; and a descending one, when each term after the first decreases by a constant ratio. Thus 2, 6, 18, 54, 162, 486 is an ascending geometrical series; and 64, 32, 16, 8, 4, 2 is a descending geometrical series. Of the former, 3 is the common ratio, and of the lat ter, 2. 564. In geometrical progression the first term, the last term, the number of terms, the common ratio, and the sum of the terms are so related to each other, that, any three of these being given, the other two may be readily determined. 565. To find any proposed term, one of the extremes, the ratio, and the number of terms being given. Ex. 1. If the first term of a geometrical series be 3, the ratio 2, and the number of terms 8, what is the last term ? Ans. 384. It is evident that the successive 3 x 2 = 3 X 128 = 384. terms are the result of repeated multiplications by the ratio ; thus the second term must be the product of the first term by the ratio, the third term the product of the second term by the ratio, and so The eighth, or last term, therefore, must be the result of seven OPERATION. on. such multiplications, or the product of the first term, 3, by 27, or 3 X 128 = 384. If the last term had been given and the first required, the process would evidently have been by division, since every less termi is the result of a division of the term next larger by the ratio. RULE. — Raise the ratio to the power whose index is one less than the number of terms ; by which multiply the least term to find the greatest, or divide the greatest to find the least. Note 1. – When the ratio requires to be raised to a high power, the process may be abridged, as in Art 516. NOTE 2. — The rule may be applied in computing compound interest, the principal being the first term, the amount of one dollar for one year the ratio, the time, in years, one less than the number of terms, and the amount the last term. EXAMPLES 2. If the first term be 5, and the ratio 3, what is the seventh term ? Ans. 3645. 3. If the series be 72, 24, 8, &c., and the number of terms 6, what is the last term ? Ans. 4. If the larger extreme be 885735, the ratio f and the number of terms 12, what are the tenth and the eleventh terms ? Ans. 45 and 15. 5. If the seventh term is 5, and the ratio }, what is the first term ? Ans. 3645. 6. If the first term is 50, the ratio 1.06, and the number of terms 5, what is the last term ? Ans. 63.123848. 7. If I were to buy 30 oxen, giving 2 cents for the first ox, 4 cents for the second, 8 cents for the third, &c., what would be the price of the last ox? Ans. $ 10737418.24. 8. What is the amount of $ 160.00 at compound interest for 6 years? Ans. $ 226.96305796096. 9. What is the amount of $ 300.00 at compound interest at 5 per cent. for 8 years ? Ans. $ 443.236+. 10. What is the amount of $ 100.00 at compound interest at 6 cent. for 30 years ? Ans. $ 574.349117291325011626410633231080264584635 7252196069357387776. per 566. To find the sum of a series, the first term, the ratio, and the number of terms being given. OPERATION. Ex. 1. If the first term be 1, the ratio 3, and the number of terms 5, required the sum of the terms. Ans. 121. If we multiply the series 1, 3, 9, 27, (81 X 3)-1 81 by the ratio 3, we shall obtain as a = 121. 3 1 second series 3, 9, 27, 81, 243, whose sum is three times the sum of the first series, and the difference between whose sum and the sum of the first series is evidently twice the sum of the first series. Now it will be observed that the two series have their terms alike, with the exception of the first term in the first series, and the last in the second series. We have then only to subtract the first term in the first series from the last term in the second, and the remainder is twice the sum of the first series; and half this being taken gives the required sum of the series. Therefore the sum of the first series must be 242 = 2 121. RULE. Find the last term as în Art. 565, multiply it by the ratio, and the procluct less the first term divide by the ratio less 1; the result will be the sum of the series. Note 1. — If the ratio is less than a unit, the product of the last term mulplied by the ratio must be subtracted from the first term, and the remainder divided by unity or 1 decreased by the ratio. NOTE 2. When a descending series is continued to infinity, it becomes what is called an infinite series, whose last term must always be regarded as 0, and its ratio as a fraction. To find the sum of an infinite series, Divide the first term by 1 decreased by the fraction denoting the ratio, and the quotient will be the sum required. This process furnishes an expeditious way of finding the value of circulating decimals, since they are composed of numbers in geometrical progression, whose common ratios are to, idő, hobo, &c. according to the number of factors contained in the repetend. Thus, .3333, &c. represents the geometrical series ð, iðó, Toồo, &c. whose first term is ló and common ratio is EXAMPLES. 2. The first term of a series is 5, the ratio {, and the number of terms 6; required the sum of the series. (3 x 5= 363; X = $38; 5 – 2 + (1 - 1) == 13166, Ans. 3. Find the value of the circulating decimal .232323, &c. 13+ (1 — Ido) = 33, Ans. 4. What is the sum of the series 4, 1, 1, 16, &c., continued to an infinite number of terms ? 4 • (1 1) 5}, Ans. 5. If the first term is 50, the ratio 1.06, and the number of terms 4, what is the sum of the series? Ans. 218.7303. 3 3 25 729 ; 6. A gentleman offered a house for sale on the following terms; that for the first door he should charge 10 cents, for the second 20 cents, for the third 40 cents, and so on in a geometrical ratio. If there were 40 doors, what was the price of the house? Ans. $ 109951162777.50. 7. If the series Ķ, ć, ik, 24, &c. were carried to infinity, what would be its sum ? Ans. 11. 8. A gentleman deposited annually $ 10 in a bank, from the ime his son was born until he was 20 years old. Required the amount of the 21 deposits at 6 per cent., compound interest, when his son was 21 years old? Ans. $ 423.92. 9. Find the value of .008197133, &c., continued to infinity. Ans. 99% 10. If a body be put in motion by a force which moves it 10 miles in the first portion of time, 9 miles in the second equal portion, and so, in the ratio of 1%, for ever, how many miles will it pass over? Ans. 100 miles. 567. To find the ratio, the extremes and number of terms being given. Ex. 1. If the extremes of a series are 3 and 192, and the number of terms 7, what is the ratio ? It has been shown (Art. 192 ; 3 = 64; $ 64 2, Ans. 565) that the last term of a geometrical series is equal to the product of the ratio by the first term raised to a power whose index is one less than the number of terms; hence the ratio must equal the root of the quotient of the last term by the first whose index is one less than the number of terms. RULE. Divide the last extreme by the first, and extract that root of the quotient whose index is one less than the number of terms. NOTE. — When the sum of the series and the extremes are given, the ratio may be found by dividing the sum of all the terms except the first, by the sum of all the terms except the last. OPERATION, EXAMPLES. 2. If the last term of a series is 1, the largest term 512, and the number of terms 10; what is the ratio ? Ans. 2. 3. If the extremes are 5 and 885735, and the sum of the series 1328600, what is the ratio ? Ans. 3. 4. What debt can be discharged in a year by monthly pay |