more, he would stand him in $300 : what was the price of 300 proof. 2. Two persons, A. and B., have both the same income; A. saves one-fifth of his yearly; but B., by spending $150 per annum more than A., at the end of 8 years finds himself 400 dollars in debt; what is their income, and what does each spend per annum? Ans. Their income is $500 per annum; also A. spends $400, and B. $550 per annum. 3. There is a fish whose head is 9 inches long, and his tail is as long as his head and half his body, and his body is as long as his head and tail what is the whole length of the fish? Ans. 6 feet. PROGRESSION Consists in two Parts-ARITHMETICAL and GEOMETRICAL. ARITHMETICAL PROGRESSION Is when a rank of numbers increase or decrease regularly, by the continual adding or subtracting of some equal num ber: As 1, 2, 3, 4, 5, 6, are in Arithmetical Progression by the continual increasing or adding of one; and 11, 9, 7, 5, 3, 1, by the continual decrease or subtraction of two. NOTE. When any even number of terms differ by Arithmetical Progression, the sum of the two extremes will be equal to the two middle numbers, or any two means equally distant from the extremes: As, 2, 4, 6, 8, 10, 12, where 6+8, the two middle numbers, are = 12 +2, the two extremes, and= 104, the two means = 14. 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. In Arithmetical Progression five things are to be observed, viz. 1. The first term. 2. The last term. 3. The number of terms. 4. The equal difference. 5. The sum of all the terms. Any three of which being given, the other two may be found. The first, second, and third terms given, to find the fifth. RULE. Multiply the sum of the two extremes by half the number of terms, or multiply half the sum of the two extremes by the whole number of terms, the product is the total of all the terms. EXAMPLES. 1. How many strokes does the hammer of a clock strike in 12 hours? 12113 then 13 X 678. Ans. 2. A man buys 17 yards of cloth, and gave for the first yard 2s. and for the last 10s. what did the 17 yards amount to ? Ans. £5 2s. 3. If 100 eggs were placed in a right line, exactly a yard asunder from one another, and the first a yard from a basket, what length of ground does that man travel who gathers up these 100 eggs singly, returning with every egg to the basket to put it in ? Ans. 5 miles, 1300 yards. The first, second, and third terms given, to find the fourth. RULE. From the second subtract the first, the remainder divided by the third, less one, gives the fourth EXAMPLES. 1. A man had 8 sons, the youngest was 4 years old, and the eldest 32, the increase in Arithmetical Progression: what was the common difference of their ages? Ans. 4. 32-4-28 then 28÷8-1-4 the common difference. 2. A man is to travel from Boston to a certain place in 12 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 58 miles: what is the daily increase, and how many miles distant is that place from Boston? Ans. 5 miles daily increase. Therefore as 3 miles is the first day's journey 3+5 8 second ditto. 8 13 third ditto, &c. The whole distance is 366 miles. The first, second, and fourth terms given, to find the third. RULE. From the second subtract the first, the remainder divide by the fourth, and to the quotient add 1, gives the third. EXAMPLES. 1. A person travelling into the country, went 3 miles the first day, and increased every day by 5 miles, till at last he went 58 miles in one day: how many days did he travel? Ans. 12. 58-3-55 then 55÷5-11 and 11+1=12 the number of days. 2. A man being asked how many sons he had, said that the youngest was 4 years old, and the eldest 32, and that he increased one in his family every 4 years, how many had he? Ans. 8. The second, third, and fourth given, to find the first. RULE. Multiply the fourth by the third, made less by 1, the product subtracted from the second gives the first. EXAMPLES. 1. A man in 10 days went from Boston to a certain town in the country, every day's journey increasing the former by 4, and the last day he went was 46 miles: what was the first? Ans. 10 miles. 4X10-1-36 then 46-36-10, the first day's journey. 2. A man takes out of his pocket at 8 several times, so many different numbers of shillings, every one exceeding the former by 6; the last 46: what was the first? Ans. 4. The second, third, and fifth given, to find the first. RULE. Divide the fifth by the third, and from the quotient subtract half the product of the fourth, multiplied by the third less 1, gives the first. EXAMPLE. A man is to receive £360 at 12 several payments, each to exceed the former by £4, and is willing to bestow the first payment on any one that can tell him what it is: what will that person have for his pains? Ans. £8. 360÷12-30 then 30 4X12-1 2 8. the first payment. The first, third, and fourth given, to find the second. RULE. Subtract the fourth from the product of the third, multiplied by the fourth; that remainder, added to the first, gives the second. EXAMPLE. What is the last number of an Arithmetical Progression, beginning at 6, and continuing by the increase of 8 to 20 places: Ans. 158. 20X8-8=152 then 152+6=158, the last number GEOMETRICAL PROGRESSION Is the increasing or decreasing of any rank of numbers by some common ratio, that is, by the continual multiplication or division of some equal number: As 2, 4, 8, 16, increase by the multiplier 2, and 16, 8, 4, 2, decrease by the divisor 2. NOTE. When any number of terms is continued in Geometrical Progression, the product of the two extremes will be equal to any two means, equally distant from the extremes: As 2, 4, 8, 16, 32, 64, where 64X2-4X32=8X16=128. When the number of terms are odd, the middle term multiplied into itself will be equal to the two extremes, or any two means, equally distant from the mean : As 2, 4, 8, 16, 32, where 2×32-4X16=8X8=64. In Geometrical Progression the same five things are to be observed, as in Arithmetical, viz. 1. The first term. 2. The last term. 3. The number of terms. 4. The equal difference or ratio. 5. The sum of all the terms. NOTE. As the last term in a long series of numbers, is very tedious to come at, by continual multiplication; therefore, for the readier finding it out, there is a series of numbers, made use of in Arithmetical Proportion, called indices, beginning with an unit, whose common difference is one; whatever number of indices you make use of, set as many numbers (in such Geometrical Proportion as is given in the question,) under them : As 1, 2, 3, 4, 5, 6 indices. 2, 4, 8, 16, 32, 64 numbers in Geometrical Proportion. But if the first term in Geometrical Proportion be different from the ratio, the indices must begin with a cipher : Aso, 1,2,3, 4, 5, 6 indices. 1,2, 4, 8, 16, 32, 64 numbers in Geometrical Proportion. When the indices begin with a cipher, the sum of the indices made choice of must be always one less than the number of terms given in the question, for 1 in the indices is over the second term, and 2 over the third, &c. Add any two of the indices together, and that sum will agree with the product of their respective terms. As in the first table of indices 2+5=7 Then in the second, 4×32-128 In any Geometrical Progression proceeding from unity, the ratio being known, to find any remote term, without producing all the intermediate terms. RULE. Find what figures of the indices, added together, would give the exponent of the term wanted, then multiply the numbers standing under such exponent into each other, and it will give the term required. NOTE. When the exponent 1 stands over the second term, the number of exponents must be one less than the number of terms. EXAMPLES. 1. A man agrees for 12 peaches, to pay only the price of the last, reckoning a farthing for the first, a half-penny for the second, &c. doubling the price to the last what must he pay for them ? |