3. A man puts out $100 at simple interest, at 7 per cent.; at the end of the 1st year it will have increased to $107, at the end of the 2d year to $114, and so on, increasing $7 each year: what will be the amount at the end of 16 years? Ans. $205. 272. Since the last term of an arithmetical progression is equal to the first term added to the product of the common difference by 1 less than the number of terms, it follows, that the difference of the extremes will be equal to this product, and that the common difference will be equal to this product divided by 1 less than the number of terms. Hence, we have CASE II. Having given the two extremes and the number of terms of an arithmetical progression, to find the common difference. Subtract the less extreme from the greater and divide the remainder by 1 less than the number of terms: the quotient will be the common difference. EXAMPLES. 1. The extremes are 4 and 104, and the number of terms 26: what is the common difference? We subtract the less extreme from the greater and divide the difference by one less than the number of terms. OPERATION. 104 4 26-1=25)100(4 2. A man has 8 sons, the youngest is 4 years old and the eldest 32, their ages increase in arithmetical progression: what is the common difference of their ages? Ans. 4. 32-4-28: then, 8-1-7)28(4. 3. A man is to travel from New York to a certain place in 12 days; to go 3 miles the first day, increasing every day by the same number of miles; so that the last day's journey may be 58 miles: required the daily increase. Ans. 5 miles 272. How do you find the common difference, when you know the two extremes and number of terms? 273. If we take any arithmetical series, as 357 9 11 13 15 17 19, &c. 19 17 15 13 11 9 7 5 3 by reversing the or22 22 22 22 22 22 22 22 22 der of the terms. Here we see that the sum of the terms of these two series is equal to 22, the sum of the extremes, multiplied by the number of terms; and consequently, the sum of either series is equal to the sum of the two extremes multiplied by half the number of terms; hence, we have CASE III. To find the sum of all the terms of an arithmetical progression, Add the extremes together and multiply their sum by half the number of terms: the product will be sum of the series. EXAMPLES. 1. The extremes are 2 and 100, and the number of terms 22: what is the sum of the series? We first add together the two`extremes, and then multiply by half the number of terms. 2. How many strokes strike in 12 hours? OPERATION. 2 1st term 100 last term 102 sum of extremes 11 half the number of terms 1122 sum of series. does the hammer of a clock Ans. 78. 3. The first term of a series is 2, the common difference 4, and the number of terms 9, what is the last term and sum of the series? Ans. last term 34, sum 162. 4. If 100 eggs are placed in a right line, exactly one yard from each other, and the first one yard from a basket: what distance will a man travel who gathers them up singly, and places them in the basket? Ans. 5 miles, 1300 yards. 273. How do you find the sum of an arithmetical series? GEOMETRICAL PROGRESSION. 274. If we take any number, as 3, and multiply it continually by any other number, as 2, we form a series of numbers: thus, 3 6 12 24 48 96 192, &c., in which each number is formed by multiplying the number before it by 2. This series may also be formed by dividing continually the largest number 192 by 2. Thus, 192 96 48 24 12 6 3. A series formed in either way, is called a Geometrical Series, or a Geometrical Progression, and the number by which we continually multiply or divide, is called the common ratio. When the series is formed by multiplying continually by the common ratio, it is called an ascending series; and when it is formed by dividing continually by the common ratio, it is called a descending series. Thus, 3 6 12 24 48 96 192 is an ascending series. 192 96 48 24 12 6 3 is a descending series. The several number are called terms of the progres sion. The first and last terms are called the extremes, and the intermediate terms are called the means. J 275. In every Geometrical, as well as in every Arithmetical Progression, there are five things which are considered, any three of which being given or known, the remaining two can be determined. They are, 274. How do you form a Geometrical Progression? What is the common ratio? What is an ascending series? What is a descending series? What are the several numbers called? What are the first and last terms called? What are the intermediate terms called? progression, how many things are conHow many must be known before the What is any term equal to? How 275. In every geometrical sidered? What are they? remaining ones can be found? do you find the last term? 1st, the first term, 2d, the last term, 5th, the sum of all the terms. By considering the manner in which the ascending progression is formed, we see that the second term is obtained by multiplying the first term by the common ratio; the 3d term by multiplying this product by the common ratio, and so on, the number of multiplications being one less than the number of terms. Thus, 3=3 1st term, 3x2=6 2d term, 3×2×2=12 3d term, 3×2×2×2=24 4th term, &c. for the other terms. But 2×2=22, 2×2×2—23, and 2×2×2×2=24. Therefore, any term of the progression is equal to the first term multiplied by the ratio raised to a power 1 less than the number of the term. CASE I. Having given the first term, the common ratio, and the number of terms, to find the last term, Raise the ratio to a power whose exponent is one less than the number of terms, and then multiply the power by the first term: the product will be the last term. EXAMPLES. 1. The first term is 3 and the ratio 2: what is the 6th term? 2×2×2×2×2=25=32 3 1st term. Ans. 96 2. A man purchased 12 pears: he was to pay 1 farthing for the 1st, 2 farthings for the 2d, 4 for the 3d, and so on doubling each time: what did he pay for the last? Ans. £2 2s. 8d. 3. A gentleman dying left nine sons, and bequeathed his estate in the following manner; to his executors £50; his youngest son to have twice as much as the executors, and each son to have double the amount of the son next younger: what was the eldest son's por. tion? Ans. £25600. 4. A man bought 12 yards of cloth, giving 3 cents for the 1st yard, 6 for the 2d, 12 for the 3d, &c.: what did he pay for the last yard? Ans. $61,44. CASE II. 276. Having given the ratio and the two extremes to find the sum of the series. Subtract the less extreme from the greater, divide the remainder by 1 less than the ratio, and to the quotient add the greater extreme: the sum will be the sum of the series. EXAMPLES. 1. The first term is 3, the ratio 2, and last term 192: what is the sum of the series? 192-3-189 difference of the extremes, 2-1-1) 189 (189; then 189+192-381 Ans. 2. A gentleman married his daughter on New Year's day, and gave her husband 1s. towards her portion, and was to double it on the first day of every month during the year: what was her portion ? Ans. £204 15s. 3. A man bought 10 bushels of wheat on the condition that he should pay 1 cent for the 1st bushel, 3 for the 2d, 9 for the 3d, and so on to the last: what did he pay for the last bushel, and for the 10 bushels? Ans. Last bushel $196,83, total cost $295,24. 4. A man has 6 children; to the 1st he gives $150, to the 2d $300, to the 3d $600, and so on, to each twice as much as the last: how much did the eldest receive, and what was the amount received by them all? Ans. Eldest $4800, total $9450, 276. How do you find the sum of the series? |