Hence, we have CASE I. Having given the first term, the common difference, and the number of terms, to find the last term. RULE. Multiply the common difference by 1 less than the number of terms, and to the product add the first term. Q. How do you find the last term when the first term and common difference are known? EXAMPLES. 1. The first term is 3, the common difference 2, and the number of terms 19: what is the last term? 2. A man bought 50 yards of cloth for which he was to pay 6 cents for the first yard, 9 cents for the 2nd, 12 cents for the 3d, and so on increasing by the common difference 3: how much did he pay for the last yard? Ans. $1,53. 3. A man puts out $100 at simple interest, at 7 per cent; at the end of the first year it will have increased to $107, at the end of the 2nd year to $114, and so on, increasing $7 each year: what will be the amount at the end of 16 years? Ans. $205. 4. Twelve persons agree to contribute to a charitable object in the following proportions: the first person is to give $2, the 2nd $4, the 3rd $6, and so op, each giving $2 more than the one previous: what does the last one give? Ans. $24. 5. The first term is 5, the common difference 12, and the numbers of terms 15: what is the last term? Ans. 173. § 191. 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. RULE. 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. Q. How do you find the common difference, when you know the two extremes and number of terms? 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. 26-1-25)100(4 100 Ans. 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? 32-4-28: then 8-1-7)28(4. Ans. 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. § 192. If 3 5 7 we take any arithmetical series, as 9 11 13 15 17 19, &c. 19 17 15 13 11 9 7 5 3 by reversing the order 22 22 22 22 22 22 22 22 22| 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. RULE. Add all the extremes together and multiply their sum by half the number of terms, the product will be the sum of the series. Q. How do you find the sum of an arithmetical 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. OPERATION. 2 1st term 100 last term 102 sum of extremes 11 half the number of terms 1122 sum of series. Ans. 1122. 2. How many strokes does the hammer of a clock strike in 12 hours? 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. GEOMETRICAL PROGRESSION. § 193. 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 96 192 192 96 48 24 12 6 2 is an ascending series. is a descending series. The several numbers are called terms of the progression. The first and last terms are called the extremes, and the intermediate terms are called the means. Q. 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? § 194. 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. 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 3rd 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=1 1st term, 3x2=6 2nd term, 3×2×2=12 3rd 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. Q. In every Geometrical Progression, how many things are considered? What are they? CASE I. Having given the first term, the common ratio, and the number of terms, to find the last term. RULE. 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 first, 2 farthings for the 2nd, 4 for the 3rd, 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; |