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the correction of the number belonging to the least error, which error is to be added or subtracted according as that number was too little or too great.
Examples. 1. A gentleman has two horses, worth $200: he has also a saddle, which, if put upon the first horse, makes his value double that of the second ; and if put upon the second horse, makes his value equal to the first-I wish to know the value of each horse, also the value of the saddle. First-Suppose $110, for the value of the first horse,
Then will 90 be the value of the second horse.
180 the value of the first horse and saddle. 110 value of the first horse.
70 value of the saddle alone.
160 value of the second horse and sad
dle, which, according to the ques
tion, is equal to the first horse. 110 value of the first horse.
50 error too little. Again-Suppose $125 for the value of the first horse. Then will 75 be the value of the second horse.
150 value of the first horse.
25 value of the saddle.
100 value of the second horse and saddle,
which, according to the question, is
equal to the first horse. 125 value of the first horse.
25 error too much.
Sup. Errors. 110..50 6250 125 25 2750
75 9000(120 Ang. Value of the first horse.
120 value of the first horse,
By Note 2.
25 less error.
75 sum of the errors.
75)375(5 correction belonging to the least error; which is to
375 be subtracted, because the number is too great. Hence 125–5 = 120 value of the first horse, as before.
2. A man had 2 silver cups, weighing together 20 ounces, and having one cover for both, now if the cover be put on the lesser cup, it will be half the weight of the greater cup; and set it on the greater cup, it will be five times as heavy as the lesser cup-What is the weight of each cup; also of the cover?
Ans. Lesser cup, 4 ounces.
Greater cup, 16
Cover, 4 3. A young gentleman having asked his father how old he was, received the following reply: 12 years ago my age was in a four-fold ratio to yours; but if we should both happen to live 6 years hence, my age will be just double to yours.—I desire to know their several ages.
Ans. 48 and 21 years. 4. A gentleman caught a fish, whose head was 8 inches long, the tail as long as the head and half the body, the body was just the length of the head and tail-What was the length of the whole fish?
Ans. 5 feet 4 inches.
5. A. B. and C. discoursing of their ages, A. affirmed that he was 22 years of age; B. said his age was equal to that of A. and half the age of C.; and C. affirmed that he was as old as both A. and B.—What was the
of each person? Ans. A. 22, B. 66, and C. 88.
ARITHMETICAL PROGRESSION, ARITHMETICAL PROGRESSION is a rank or series of numbers which increase or decrease regularly, by a common difference, that is, by the continual adding or subtracting of an equal number.
In arithmetical progression five things are to be observed, viz.
1. The first term.
Any three of which being given, the rest may be found. Note-When any even number of terms differ, by arithmetical progression, the sum of the two extremes will be equal to the sum of the two middle numbers, or any two means equally distant from the extremes; as 3, 5, 7, 9, 11, 13; where 7+9, the two middle numbers, are = 3+13, the two extremes, and
5+11, the two means 16. When the number of terms are odd, the double of the middle term will be equal to the two extremes, or of any two means equally distant from the middle term; as 1, 2, 3, 4, 5, where the double of 3 = 5+1 = 2+4: 6.
CASE I. The first term, common difference, and number of terms given, to find the last term, and sum of all the terms.
RULE. Subtract the common difference from the product of the number of terms, multiplied by the common difference, the remainder, added to the first term, gives the last term. Multiply the sum of the two extremes, (the first and last
terms) by the number of terms, and half the product will be the sum of the series.
Examples. 1. Twenty-five persons bestowed charity to a poor man; the first gave him 10 cents, the second 12, and so on in arithmetical progression-What did the last person give, and what sum did the man receive ?
25 number of terms.
58 last term, or last person gave.
$8.50 sum received. 2. A. covenanted with B. to serve him 12 years, and to have $10 the first year, and his wages to increase annually $3, during the term—What had he the last year; what on an average yearly, and what for the whole time?
843 the last year. Ans. 26.50 annually.
318 the whole time,
CASE II. When the two extremes and number of terms are given, to find the common difference.
RULE. Divide the difference of the extremes by the number of terms less one, the quotient will be the common difference.
Examples. 1. Admit a debt be discharged at 14 several payments, in arithmetical progression, the first to be 820, the last 898 What is the common difference, and what the whole debt?
14—1=13)78(6 common difference.
$826 whole debt.
When a number of quantities increase by the same multiplier, or decrease by the same divisor, they form a Geometrical Series. . This common multiplier, or divisor, is called the ratio. Thus 2, 4, 8, 16, 32, 64, &c. increase by the continual multiplication of 2; and 64, 32, 16, 8, 4, 2, decrease continually by the division of 2, or multiplication of .5.
RULE. Multiply the first term into such a power of the ratio as is indicated by the number of terms less one, and the product will be the last term.
Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio less' one; the quotient will be the sum of the series.
Examples. 1. If a man were to engage to pay another 1 cent for the first month, 6 for the second, 36 for the third, and so on, in a six-fold ratio, for 12 months' service. How much would his wages amount to?