CASE I. Having given the first term, the common difference, and the number of terms, to find the last term. Multiply the common difference by 1 less than the number of terms, and to the product add the first term. EXAMPLES. 1. The first term is 3, the common difference 2, and the number of terms 19: what is the last term? We multiply the number of terms less 1, by the common difference 2, and then add the first term. OPERATION. 18 number of terms less 1. 2 common difference 36 3 1st term. 39 last term. 2. A man bought 50 yards of cloth; he was to pay 6 cents for the first yard, 9 cents for the 2d, 12 cents for the 3d, and so on increasing by the common difference 3: how much did he pay for the last yard? 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 2d year to $114, and so on, increasing $7 each year: what will be the amount at the end of 16 years? 319. 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 re QUEST.-319. How do you find the common difference, when you know the two extremes and number of terms? mainder 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? 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? 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. 320. If we take any arithmetical series, as &c. by reversing the order of the terms. 3 5 7 9 11 13 15 17 19, 19 17 15 13 11 9 7 5 3 22 22 22 22 22 22 22 22 22 | 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 gression, pro 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? QUEST.-320. How do you find the sum of an arithmetical 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. 2. How many times does the hammer of a clock strike in 12 hours? 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? 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? GENERAL EXAMPLES. 1. What is the 18th term of an arithmetical progression of which the first term is 4 and the common difference 5? 2. The 18th term of an arithmetical progression is 89 and the common difference 5: what is the first term? 3. A flight of stairs has 18 steps; the first ascends but 12 inches in a vertical line, and each of the others 18: what is the entire ascent in a vertical line? 4. A debtor has 18 creditors; he owes to the largest creditor 89 dollars, and 5 dollars less to each of the others in succession: how much does he owe to the least? 5. A person travelled from Boston to a certain place in 8 days; he travelled 2 miles the first day, and every succeeding day he travelled farther than he did the preceding by an equal number of miles: the last day he travelled 23 miles: how much did he travel each day, and how much in all? 6. The number of terms is 22, the common difference 5, and the sum of the terms 1221: what is the least term? 7. A man is to receive $3000 in 12 payments, each succeeding payment to exceed the previous by $4: what will the last payment be? GEOMETRICAL PROGRESSION 321. 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 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 numbers are called terms of the progression. The first and last terms are called the extremes, and the intermediate terms are called the means. 322. 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, QUEST.-321. 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? 322. Ir every geometrical progression, how many things are considered? What are they? 1st, the first term, 2d, the last term, 3d, the common ratio, 4th, the number of terms, 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 31st term, 3 × 2 = 6 2d term, 3 × 2 × 2 = 12 3d term, 3 × 2 × 2 × 2 24 4th term, &c. for the other terms. But 2 × 2 = 23, 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 — 23 : 25 = 32 3 1st term. Ans. 96 QUEST.-How many must be known before the remaining ones can be found? What is any term equal to? How do you find the last term? |