RULE.-Divide the sum of the series by the number of terms; from this quotient, subtract half the product of the number of terms less one by the common difference, and the remainder will be the least term. EXAMPLES. 1. The number of terms is 10, their sum 155, and common difference 3. Required the least term of the series. Here 155 X 32 term. 10 = 15; and (101) x 32=9 27 ÷ 2 = 13; then 15 13 = 2 least 2. The number of terms is 36, their sum 405, and the common difference. Required the least term of the series. Ans. 24. 3. A man travels 10100 miles in 100 days; the first day he travels a certain number of miles, and each succeeding day two miles more. How many miles did he travel the first Ans. 2 miles. day? 4. A young lady is to receive £558 2s. 11d. in one year, for her portion, by successive daily payments, each succeeding payment to exceed the former by 2d. What will her first payment be? Ans. 3d. PROBLEM V. Given the least term, the common difference, and the number of terms to find the greatest term. RULE.-Multiply the number of terms by the common difference, to this product add the least term; from this sum subtract the common difference, and the remainder will be the greatest term. EXAMPLES. 1. The least term is 3, the common difference 2, and the number of terms 365. Required the greatest term. Here (365 × 2) + 3 = 730 +3 733; then 733 2731 Ans. 2. The least term is 4, the common difference 4, and the number of terms 300. Required the greatest term. Ans. 1200, 4. Bought 100 yards of cloth, for the first yard I paid 3d. for the second 7d, and so on, for each succeeding yard. Required the price of the last yard, and the value of the whole. Ans. £1. 13s. 3d. last yard, and £83 15s. for the whole. PROBLEM VI.-Given the number of terms, their common difference, and the greatest term, to find the least term. RULE.-Multiply the number of terms less 1, by the common difference; and subtract this product from the greatest term, and the remainder will be the least term. EXAMPLES. 1. The greatest term in a series of numbers in arithmetical progression is 23, the number of terms 8, and the common difference 3. Required the least term. Here (81) × 3 = 7 × 3 = 21; 2 Ans. and 23 21 = 2. The greatest term is 20, the number of terms 36, and the common difference . Required the least term, Ans. 2. 3. A person who had a daily allowance for one year, each day's pay being 2d. more than the preceding, received £3. 11d. for the last day. How much was paid for the first day, and what sum for the year ? Ans. 3d. for the first day, £558 2s. 11d. for the year. GEOMETRICAL PROGRESSION. I GEOMETRICAL PROGRESSION is a series of numbers or quantities which increase or decrease by a common ratio, as 2, 4, 8, 16, 32, 64, &c., and 81, 27, 9, 3, 1, 1, 1, 27, &c., are series in geometrical progression; the ratio of the first being 2, and that of the second 3. The first is called an increasing series, and the second a decreasing series. The first and last terms are called the extremes, and the common multiplier or divisor the ratio. The middle term is called the mean, when the numbers are odd; but when even, the two middle terms are called means. If three numbers be in geometrical progression the product of the extremes is equal to the square of the mean. Thus, if the geometrical series 4, 8, 16, be taken; 4 × 16 = 8 x 8. If four numbers, in geometrical progression, be taken, as 4, 8, 16, 32; Then 4 X 32 = 8 X 16. If the series 2, 4, 8, 16, 32, 64, 128 be taken ; Then 2 × 128 = 16 × 16. If the series 81, 27, 9, 3, 1, †, †, 27 be taken; Or, the product of the two extremes will be equal to the product of any two means, equidistant from the extremes. Then 2 × 128 = 4 × 64 ; 0r 81 × 7 = 9 x }. The equidistant terms, of a series of numbers in geometrical progression, are in geometrical progression. Thus, in the series 2, 4, 8, 16, 32, 64, 128, &c., whose ratio is 2. Then 4, 16, 64, and the equidistant terms, are in geometrical progression, whose ratio is 4. PROBLEM I. Given the first term, the ratio, and the number of terms to find the last or any other assigned term. RULE -Find such a power of the ratio as is denoted by the number of terms less one, multiply this result by the first term, and the product will be the last term. NOTE. If the first term and ratio be the same, find such a power of the ratio as is denoted by the number of terms, and the resuit will be the last term. EXAMPLES. 1. The first term of a series of numbers in geometrical progression is 2, the ratio 3, and the number of terms 12. Required the last term. Here 12 1= -- 11 the number of terms less one; Then the ratio 3 raised to the 11th. power as follows: Hero 3" = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 = 177147 the 11th. power of the ratio. 17. Then 177147 × 2 = 354294 the last term. 2. The first term is 5, the ratio 3, and the number of terms Required the last term of the series. Ans. 215233605. 3. The first term of a decreasing series in geometrical progression is 12, the ratio or divisor 2, and the number of terms 9. Required the last term. Ans. 34 4. Bought 25 yards of cloth, the first yard cost 2d., the second 4d., the third 8d., &c. What was the value of the last yard? Ans. £139810 2s. 8d. 5. A thrasher worked 30 days, for the first day's work he received 4 wheat corns, for the second 12, and for the third, 36, &c. What did he receive for his last day's work; if 8000 grains of wheat make a pint, and the wheat be worth 8s. per bushel? Ans. £214469929 5s. 3 d. 6. A person wishes to dispose of a horse for the value of the last nail in his shoes, on the following terms, viz. one farthing for the first nail, two farthings for the second, one penny for the third, and so on; admiting the 4 shoes to contain 32 nails. What would the horse be sold for? Ans. £2236962 2s. 8d. PROBLEM II.-Given the extremes and the ratio of a series of numbers in geometrical progression, to find their sum. RULE.-Multiply the greatest term of the series by the ratio, and divide the difference of this product andthe least term, by the difference between one and the ratio, and the quotient will be the sum required. EXAMPLES. 1. The first term is 2, the last term 486, and the ratio 3. Required the sum of the series. Here (486 x 3) − 2 = 1458 2 = 1456, and 3 = 2; then 14562728 Ans. 1 2. The first term of a decreasing geometrical series is 162, the last term, and the ratio 3. Required the sum. Here 162 × 3 = 486 -릉= 4857; and 3 2186 4372 ÷ 2 = = 9 1= 2; then 4857 ÷ 2 = 2428 Ans. 3. The first term is 3, the last term 6144, and the ratio What is the sum of the series? 2. 4. The first term is 14, the last term 295244, What is the sum of the series? 5. The first term is 100, the last term, What is the sum of the series? 6. Required the value of the 25 In the 4th. question, problem 1st. Ans. 12285. and the ratio 3. Ans. 44286. and the ratio 2. Ans. 19912. yards of cloth, mentioned Ans. £279620 5s. 2d. 7. How much money would the thrasher, in the fifth question, problem first, receive for his 30 days' wages? Ans. £321704893 17s. 11 d. nearly. 8. What would be the value of a horse on the following terms, viz. one farthing being paid for the first nail in the shoe, 2 farthings for the second, a penny for the third, and so on; admitting the 4 shoes to contain 9 nails each ? Ans. £71582788 5s. 3 d. 9. One Sessa, an Indian, having first discovered the game of chess, showed it to his prince, Shehram, who was so delighted with the invention, that he bid him ask what he would, as a reward for his ingenuity: upon which, Sessa requested that he might be allowed one grain of wheat for the first square on the chess-board, two for the second, four for the third, and so on, doubling continually to 64, the whole number of squares: now, supposing a pint to contain 8000 grains; it is required to find what number of ships, each carrying 1000 tons, might be freighted with the produce, allowing 40 bushels to a ton? Ans. 900719925.474099199951171875 ships. ANNUITIES. The term Annuity is understood to signify any interest of money, rent, or pension, payable from time to time, at stated periods; and these payment may take place yearly, halfyearly, quarterly, &c, Annuities are of two kinds, namely, possession and reversion. Annuities in possession are such as are entered upon, and annuities in reversion are such as cannot be entered upon, till some future period or event takes place. An annuity is in arrears, when it is not paid for any certain period after it becomes due; and the sum of all the single payments, together with the interest due thereon, is the amount. N.B. When no period of payment is stated, yearly payments are understood. When an annuitant disposes of his annuity, the sum paid for it is the present worth. |