4 x 35 x 3 - 4 3-1 SOLUTION. We have all the conditions of 460, except that the last term is needed. Since the first term, the ratio, and number of terms are given, we find the last term by 459. The problem is then similar to 460, and is solved in the same way. RULE. Raise the ratio to a power indicated by the number of terms, and subtract 1 from the result; then multiply this remainder by the first term, and divide the product by the ratio less 1. 2. The first term is 7, the ratio 3, and the number of terms 4. What is the sum of the series? Ans. 280. 3. The first term is 375, the ratio, and the number of terms 4. What is the sum of the series? 4. The first term is 175, the ratio 1.06, and the number of terms 5. What is the sum of the series? Ans. 986.49+. 5. The first term is 4, the ratio 5, the number of terms 5. What is the sum of the series? 6. What yearly debts can be discharged by monthly payments, the first being $2, the second $6, and the third $18, and so on in geometrical progression? Ans. $531440. 7. If a grain of wheat produces 7 grains, and these are sown the second year, each yielding the same increase, how many bushels will be produced at this rate in 12 years, if 1000 grains make a pint? Ans. 252315 bu. 4 qt. 8. Six persons of the Morse family came to this country 200 years ago. Suppose that their number has doubled every 20 years since, what would it be now? COMPOUND INTEREST BY GEOMETRICAL PROGRESSION. 462. We have seen (318) that if any sum at compound interest is multiplied by the amount of $1 for the given interval, the product will be the amount of the given sum or principal at the end of the first interval; and that this amount constitutes a new principal for the second interval, and so on for a third, fourth, or any other interval. Hence, 463. A question in compound interest constitutes. a geometrical progression, whose first term is the principal; the common multiplier or ratio is 1 plus the rate per cent for one interval; the number of terms is equal to the number of intervals + 1; and the last term is the amount of the given principal for the given time. All the usual cases of compound interest, and discount computed at compound interest, can therefore be solved by the rules for geometrical progression. EXAMPLES. 464. 1. Find the amount of $250 for 4 years, at 6% compound interest. $250 x 1.064: = OPERATION. $250 x 1.262477 = $316.21925. SOLUTION. Here we have $250, the first term, 1.06, the ratio, and 5, the number of terms, to find the last term. Then by 459 we find the last term, which is the amount required. 2. What is the amount of $350 in 4 years, at 6% per annum, compound interest? Ans. $441.86. 3. Of what principal is $150 the compound interest for 2 years, at 7% ? 4. In how many years will $40 amount to $53.24, at 10% compound interest? Ans. 3 years. ANNUITIES. 465. An Annuity is literally a sum of money which is payable annually. The term is, however, applied to a sum which is payable at any equal intervals, as monthly, quarterly, semi-annually, etc. Annuities are of three kinds: Certain, Contingent, and Perpetual. NOTE. The term interval will be used to denote the time between payments. 466. A Certain Annuity is one whose period of continuance is definite or fixed. 467. A Contingent Annuity is one whose time. of commencement, or ending, or both, is uncertain ; and hence the period of its continuance is uncertain. 468. A Perpetual Annuity or Perpetuity is one which continues forever. 469. Each of these kinds is subject, in reference to its commencement, to the following conditions: 1. It may be deferred, i.e. it is not to be entered upon until after a certain period of time. 2. It may be reversionary, i.e. it is not to be entered upon until after the death of a certain person, or the occurrence of some certain event. 3. It may be in possession, i.e. it is to be entered upon at once. 470. An Annuity in Arrears or Forborne is one on which the payments were not made when due. Interest is to be reckoned on each payment of an annuity in arrears, from its maturity, the same as on any other debt. ANNUITIES AT SIMPLE INTEREST. 471. In reference to an annuity at simple interest, we observe: 1. The first payment becomes due at the end of the first interval, and hence will bear interest until the annuity is settled. 2. The second payment becomes due at the end of the second interval, and hence will bear interest for one interval less than the first payment. 3. The third payment will bear interest for one interval less than the second; and so on to any number of terms. Hence, 4. All the payments being settled at one time, each will be less than the preceding, by the interest on the annuity for one interval. Therefore, they will constitute a descending arithmetical progression, whose first term is the annuity plus its interest for as many intervals less one as intervene between the commencement and settlement of the annuity; the common difference is the interest on the annuity for one interval; the number of terms is the number of intervals between the commencement and settlement of the annuity; and the last term is the annuity itself. 472. The rules in Arithmetical Progression will solve all problems in annuities at simple interest. EXAMPLES. 1. A man works for a farmer one year and six months, at $20 per month, payable monthly; and these wages remain unpaid until the expiration of the whole term of service. How much is due to the workman, allowing simple interest at 6% per annum ? OPERATION. $20 + $.10 x 17 = $21.70, 1st term. × 18 = $375.30, sum. SOLUTION.- Here the last month's wages, $20, is the last term; the number of months, 18, is the number of terms; and the interest on 1 month's wages, $.10, is the common difference; and since the first month's wages has been on interest 17 months, the progression is a descending series. Then, by 451, we find the first term, which is the amount of the first month's wages for 17 months; and by 454 we find the sum of the series, which is the sum of all the wages and interest. 2. A father deposits annually for the benefit of his son, commencing with his tenth birthday, such a sum that on his 21st birthday the first deposit at simple interest amounts to $210, and the sum due his son to $1860. How much is the deposit, and at what rate per cent is it deposited ? OPERATION. $1860 x 2- $210 x 12 SOLUTION. 12 210-100 11 = $100, deposit. Here the $210, the amount of the first deposit, is the first term; 12, the number of deposits, is the number of terms; and $1860, the amount of all the deposits and interests, is the sum of the series. By 454, we find the last term to be $100, which is the annual deposit; and by 452 we find the common difference to be $10, which is the annual rate per cent. 3. What is the amount of an annuity of $150 for 51 years, payable quarterly, at 11% per quarter? Ans. $3819.75. 4. In what time will an annual pension of $500 amount to $3450, at 6% simple interest? Ans. 6 years. 5. Find the rate per cent at which an annuity of $ 6000 will amount to $59760 in 8 years, at simple interest? Ans. 7%. 6. What is the present worth of an annuity of $300 for 5 years, at 6% ? 7. What is the present worth of an annuity of $500 for 10 years, at 10% ? |