5. The extremes of an arithmetical series being 6 and 76, and the number of terms 15, what is the common difference? 6. A man has 5 sons; the youngest is 6 years old, and the eldest 18; their ages increase in arithmetical progression; what is the common difference of their ages? CASE III. To find the sum of all the terms of an arithmetical series. 7. What is the sum of all the terms of the series 2, 5, 8, 11, 14, 17? 2, 5, 8, 11, 14, 17 17, 14, 11, 8, 5, 2 19, 19, 19, 19, 19, 19 If we write the series twice, reversing the order of the terms as in the margin, and add the corresponding terms, we see that twice the sum of all the terms of the series is equal to 19 multiplied by 6; or, to the product of the sum of the extremes multiplied by the number of Hence the following terms. RULE. Find the last term as in Case I., and then multiply the sum of the extremes by half the number of terms. Or, Multiply the sum of the extremes by the number of terms, and take the product. 8. How many times does the hammer of a clock strike in 12 hours? SOLUTION. (1+12)×6=78. Or, (1+12) X 12 78. 9. How many times would it strike in 24 hours, if the hours were numbered from 1 to 24? 10. How far will a stone fall in 3 seconds? in 6 seconds? in 8 seconds? NOTE. The last term is the distance it will fall in the 3d, the 6th, or the 8th second. See question 3. 11. If a young man, at the age of 21, "a temperate drinker," spends 5 cents per day, for his favorite beverage, for 300 days in the year, how much does it cost him in one year? 12. If for each succeeding year it costs him $3 more than the preceding, (a very moderate estimate,) how much will he have spent at the age of 40, if he should escape the drunkard's grave till that age ? NOTE. The last term is 3X (19—1)+15. 13. Thirteen wheels revolve in arithmetical progression; the first makes 3 and the last 51 revolutions per second. How many revolutions does each wheel exceed the former, and how many do they all together make in one second? NOTE. Find the common difference, by Case II. 14. Eleven wheels revolve in arithmetical progression. The first makes 3 revolutions, and the last 33, per second. Required the sum of the series. 148. ANNUITIES AT SIMPLE INTEREST. A sum of money due annually, quarterly, or at any other regular periods, is called an annuity. The periodical sums are sometimes called instalments. The present worth of an annuity is that sum which being put at interest for the time, will be sufficient to pay it. The amount of an annuity is the interest of all the instalments added to their sum. An annuity at simple interest is an example of arithmetical progression; in which the first term is the periodical sum, the common difference is the interest on the periodical sum for the time that elapses between two successive instalments, and the number of terms the number of instalments. 1. A man hired a house for $200 per annum, agreeing to pay the rent quarterly; but for 2 years the rent has remained unpaid. How much does he now owe, reckoning simple interest at 8 per cent. per annum on each quarterly payment? NOTE 1. The quarter's rent just due is $50. ago is $50, and the interest on it for 3 months; ago is $50, and 6 months' interest. If, then, we term in the series, what will be the ference? The number of terms? the terms? second term? The last term? That due 3 months that due 6 months call $50 the first The common difThe sum of all 2. A man bought a house-lot for $500, agreeing to pay for it in 5 equal yearly instalments, without interest. What sum, at the end of 5 years, will pay the whole amount, reckoning interest on each instalment from the time it becomes due? NOTE. The payment last due is $100 without interest; the last payment but one is $100 and one year's interest; the last but two is $100 and two years' interest, &c. If, then, we call $100 the first term, what will be the common difference? The number of terms? The last term? The sum of all the terms? From these examples, To find the amount of an annuity at simple interest, we derive the following RULE. Find the last term of the series as in Case I., Art. 147, and then the sum of the series as in Case III. 3. What is the amount of an annuity of $150, that has been in arrears, that is, that has not been paid, for 10 years, reckoning interest at 6 per cent. per annum? 4. What will an annuity of $375 amount to in 5 years, reckoning simple interest at 8 per cent.? 5. If a pension, payable quarterly, of $100 per annum, remains unpaid for 3 years, what will be due at the end of that time, reckoning simple interest at the rate of 5 per cent. per annum ? 149. GEOMETRICAL PROGRESSION. Geometrical Progression, or Series by Quotient, is a series of numbers which increase or decrease so that the quotient of any term divided by the preceding one shall give the same quotient throughout the series. This quotient is called the Common Ratio. When the ratio is more than one, the series is called an Ascending Series; when the ratio is less than one, the series is called a Descending Series. 8 Thus, in the ascending series 3, 12, 48, 192, &c., the ratio is 4. In the descending series 72, 42, 8, 23, &, the ratio is . In Geometrical Progression, 5 things are to be considered, viz.: 1. The first term. 2. The last term. 3. The common ratio. 4. The number of terms. 5. The sum of all the terms. CASE I. 1. If the first term of a geometrical progression is 5, and the ratio 3, what is the 5th term of the series? 5X3, ×3×3×35 × 3* 405. Hence The first term, common ratio, and number of terms being given, to find the last term. RULE. Raise the ratio to a power whose exponent is one less than the number of terms, and multiply the power by the first term; the product will be the last term. 2. A man bought 10 yards of cloth, for which he agreed to pay 1 cent for the first yard, 3 cents for the 2d, 9 for the 3d, &c. What did he pay for the last yard? 3. What will $20 amount to, in 8 years, at 6 per cent., compound interest? NOTE. The first term is the principal, the ratio 1.06, the number of terms 9. 4. What will $1050 amount to, in 10 years, at 5 per cent., compound interest? 5. What is the 8th term of the series 20, 60, 180, &c.? CASE II. The ratio and the two extremes being given, to find the sum of the series. 6. What is the sum of the series 5, 15, 45, 135, 405? The series 15, 45, 135, 405, 1215, is obtained by multiplying the given series by the ratio, 3. If, therefore, we subtract the first series from the 2d, the remainder will be twice the first series. From the order in which the terms are arranged, it will be seen that this remainder is obtained by subtracting the first term in the given series from three times the last term. Therefore, 1215—5—1210 is twice the sum of the given series; and 1210-605, is the sum of the given series. Hence, to find the sum of a series by quotient, we have the following RULE. Find the last term as in the preceding article; multiply the last term by the ratio, and divide the difference between the product and the first term by the difference between the ratio and 1. 7. What is the sum of 10 terms of the series 1, 4, 16, 64, &c.? 8. A man bought a house, agreeing to pay for it in 12 monthly payments. The first payment was to be $1, the 2d $3, the 3d $9, &c. How much was the last payment? How much did they all amount to? 150. ANNUITIES AT COMPOUND INTEREST. 1. What will an annuity of $1 amount to in 5 years, at 6 per cent., compound interest? This question forms a geometrical series of 5 terms, in which $1 is the first term, the amount of $1 for 1 year, or $1.06, the ratio, and the amount of $1 for 4 years the last term. There fore, 1X (1.06) the last term; (1 X 1.065) — 1 1.06-1 = = the amount. 1 X (1.06) X 1.06) — 1 1.06-1 Hence, to find the amount of an annuity at compound interest, we have this RULE. Find the sum of the series, as in the last article.* 2. What will an annuity of $50 amount to in 10 years, at 5 per cent., compound interest? NOTE. First find the amount of an annuity of $1, as in the last (1X1.051o)-1 example. X 50= = 1.05-1 3. James Phillips was born Aug. 1, 1824. Aug. 1, 1834, his father deposited for James, in the savings bank, $10, and did the same on each returning birth-day, till his son was 21 years old. What did all the deposites amount to at that time, at 6 per cent.? at 5 per cent.? 4." What would be the difference, at the end of 10 years, between two young men, A and B? A spends $100 a year, in theatres, amusements, &c., and B invests the same sum in business, in such a way that the principal and interest of one year yield him 15 per cent. for the next year?". -WAYLAND. 5. What is the difference between the amount of an annual deposite of $100 for 10 years, bearing an annual compound interest at 5 per cent., and a semi-annual deposite of $50 for * See table III. page 262. |