EXAMPLES. 1. The extremes are 3 and 29, and the number ni lerme 14, what is the common difference? 29 Extremes. Number of terms less l=13)26(2 Ans. 2. A may had 9 sons, whose several ages differed alike, the youngest was three years old, and the oldest 35; what was the common difference of their ages ? Ans. 4 years. 3. A man is to travel from New-London to a certain place in 9 days, and to go but 3 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 43 miles: Required the daily increase, and the length of the whole journey ? Ans. The daily increase is 5, and the whole journey 207 miles. 4. A debt is to be discharged at 16 different payments (in arithmetical progression,) the first payment is to be 141. the last 1001. ; What is the common difference, and the sum of the whole debt? Ans. 5l. 14s. 8d. connon difference, and 9121. the whole debt. PROBLEM III. Given the first term, last term, and cominon difference, to find the number of terms. Rulb.--Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms. EXAMPLES. 1. If the extremes be 3 and 45, and the common difference 2; what is the number of terms? Ans. 22. 2. A man going a journey, travelled the first day five iniles, the last day 45 miles, and each day increased his journey by 4 miles ; how many days did he travel, and how far ? Ans. 11 days, and the whole distance travelled 275 milcs. GEOMETRICAL PROGRESSION, IS when any rank or series of numbers increase by one common multiplier, or decrease by one common divisor; as, 1, 2, 4, 8, 16, &c. increase by the multiplier 2 ; and 27, 9, 3, 1, decrease by the divisor 3. PROBLEM I. The first term, the last term (or the extremes) and the ratio given, to find the sum of the series. RULE. Multiply the last term by the ratio, and from the product subtract the first term ; then divide the remainder by the ratio, less by 1, and the quotient will be the sum of all the terms. EXAMPLES. 1. If the series be 2, 6, 18, 54, 162, 486, 1458, and the 1 tatio 3, what is its sum total ? 3 x 1458-2 -2186 the Answer. 3-1 2. The extremes of a geometrical series are 1 and 65536, and the ratio 4; what is the sum of the series? Ans. 87381. PROBLEM II. Given the first term, and the ratio, to find any other term assigned.* CASE I. * As the last term in a long series of numbers is very tedious to be found by continual multiplications, it will be necessary for the readier finding it out, to have a series of numbers in arithmetical proportion, called indices, whɔsé common difference is 1. | When the first term of the series and the ratic are equal, the indices muist begin with the unit, and in this case, the product of any two terms is equal to that term, signified by the sum of their indices : 1. Write down a few of the leading terms of the series, and place their indices over them, beginning the indices with a unit or 1. 2. Add together such indices, whose sum shall make up the entire index to the sum required. 3. Multiply the terms of the geometrical series belonging to those indices together, and the product will be the term sought. EXAMPLES. 1. If the first be 2, and the ratio 2; what is the 13th term ? 1, 2, 3, 4, 5, indices. Then 5+5+3=13. 2, 4, 8, 16, 32, leading terms. 32 x 32 x 8=8192 Ans. 2. A draper sold 20 yards of superfine cloth, the first yard for 3d., the second for 9d., the third for 27d., &c. in triple proportion geometrical; what did the cloth come to at that rate ? The 20th, or last term, is 3486784401d. Then 3 3486784401-3 -5230176600d. the sum of all 3_1 the terms (by Prob. I.) equal to £21792402, 10s. 3. A rich miser thought 20 guineas a price too much for 12 fine horses, but agreed to give 4 cts. for the first, 16 cts. for the second, and 64 cents for the third horse, and so on in quadruple or fourfold proportion to the last : what did they come to at that rate, and how much did they cost per head one with another ? Ans. The 12 horses came to $223696, 20 cts., and the average price was $18641, 35 cts. per head. Tlius, 1 2 3 4 5, &c. indices or arithmetical series 2 4 8 16 32, &c. geometrical series. 3-4-2 the index of the fifth term, and 4x8 the fifth term. Now, 32 = CASE Il. When the first term of the series and the ratio are diffe. rent, that is, when the first term is either greater or less than the ratio.* 1. Write down a few of the leading terms of the series, and begin the indices with a cipher: Thus, 0, 1, 2, 3, &c 2. Add together the most convenient indices to make an index less by 1 than the number expressing the place of the terms sought. 3. Multiply the terms of the geometrical series together belonging to those indices, and make the product a dividend 4. Raise the first term to a power whose index is one jens than the number of the terms multiplied, and make the result a divisor. 5. Divide, and the quotient is the term sought. EXAMPLES. 4. If the first of a geometrical series be 4, and the ratio 2, what is the 7ih term ? 0, 1, 2, 3, Indices, 3+2+1=6, the index of the 7th term. -32916 the 7th term required. 16 Here the number of terms multiplied are three; therefore the first term raised to a power less than three, is the 2w power or square of 4=16 the divisor. * When the first term of the series and the ratio are different, the indices, must begin with a cipher, and the sum of the indices made choice of must be one less than the number of 'erms given in the question : because I in the indices stands over the second-term, and 2 in the indices over the third term, &c. and in this case, the product of any two terms, divided by the first is equal to that term beyond the first, signified by the sum of their indices. Thus, 0, 1, 2, 3, 4, &c. Indices. 1, 3, 9, 27, 81, &c. Geometrical series. Here 4 : 3-7 the index of the Mil, tern. 81 x 27=2137 the 8th term, or the 7th beyond the 1st. 1 5. A Goldsmith sold 1 lb. of gold, at 2 cts. for the first ounce, 8 cents for the second, 32 cents for the third, &c. in a quadruple proportion geometrically: what did the whole come to ? Ans. $111848, 10 cts. 6. What debt can be discharged in a year, by paying 1 farthing the first month, 10 farthings, or (23d) the second, and so on, each month in a tenfold proportion? Ans. £115740740 14s. 9d. 3 qrs. 7. A thrasher worked 20 days for a farmer, and received for the first days work four barley-corns, for the second 12 barley corns, for the third 36 barley corns, and so on, in triple proportion geometrically. I demand what the 20 day's labour came to supposing a pint of barley to contain 7680 corns, and the whole quantity to be sold at 2s. 6d. per bushel ? Ans, £1773 7s. 6d. rejecting remainders. 8. A man bought a horse, and by agreement, was to give a farthing for the first nail, two for the second, four for the third, &c. 'There were four shoes, and eight nails in each shoe ; what did the horse come to at that rate? Ans. £4473924 58. 33d. 9. Suppose a certain body, put in motion, should move the length of 1 barley-corn the first second of time, one inch the second, and three inches the third second of time, and so continue to increase its motion in triple proportion geometrical ; how many yards would the said body move in the term of half a minute. Ans. 953199685623 yds. I ft. 1 in. 16. which is no less than five hundred and forty-one millions of miles. 1 POSITION. POSITION is a rule which, by false or supposed numbers, taken at pleasure, discovers the true ones required.-It is divided into two parts, Single or Double. SINGLE POSITION IS when one number is required, the properties of which are given in the question. |