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2. If the first term of a series be 8, the last 108, and the number of terms 21, what is the common difference? 108-8-21-1=5 Ans.

3. A man has 12 sons whose ages are in arithmetical progression, the youngest is 2 years old, and the oldest 35; what is the common difference

in their ages? Ans. 3 y'rs.

278. If I give 3 cents for the first lemon, and 11 cents for the last, and the common difference in the prices be 2 cents, how many did I buy?

The difference of the extremes divided by the number of terms, less 1. gives the common difference (277); consequently the difference of the extremes divided by the common difference must give the number of terms less 1, (11-3-8, and 82-4 and 4+1=) 5 Aus. Hence,

III. The first term, the last term, and the common difference given to find the number of terms.

RULE. Divide the difference of the extremes by the common difference, and the quotient, increased by 1, will be the answer.

2. If the first term of a series be 8, the last 108, and the common difference 5, what is the number of terms? 108-8-5-20, and 20+1=1 21 Ans.

3. A man on a journey travelled the first day 5 miles, the last day 35 miles, and increased his travel each day by 3 miles, how many days did be travel?

Ans. 11.

279. If I buy 5 lemons, whose prices are in arithmetical progression, giving for the first 3 cents, and for the last 11 cents, what do I give for the whole?

The mean, or average price of the lemons will obviously be half way between 3 and 11 cents the difference between 3 and 11 added to 3 is (11-32)7, and 7 the mean price multiplied by 5 the number of lemons equals (75) 35 cents the answer. Therefore,

IV. The first and last term, and the number of terms given to find the sum of the series.

RULE.-Multiply half the sum of the extremes by the number of terms, and the product will be the sum of the series.

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2. Geometrical Progression.

280. A Geometrical Progression is a series of terms which increase by a constant multiplier, or decrease by a constant divisor, as 2, 4, 8, 16. 32, &c. increasing by the constant multiplier 2,or 27, 9, 3,1, 3, &c. decrea sing by the constant divisor, 3. The multiplier or divisor, by which the series is produced, is called the ratio.

281. A person bought 6 brooms, giving 3 cents for the first, 6 cents for the second, 12 for the third, and so on, doubling the price to the sixth; what was the price of the sixth? or, in other words, if the first term of a series be 3, the number of terms 6, and the ratio 2, what is the last term? The first term is 3, the second, 3x2=6, the third, 6×2=(3×2×2)= 12, the fourth, 12×2=(3×2×22) 24, the fifth, 2423×2×2×2 2) 48, and the sixth, 482=(3×2×2×2×2×2) 96. Then 96 cents is the cost of the sixth broom. By examining the above, it will be seen that the ratio is, in the production of each term of the series, as many times a factor, less one, as the number of terms, and that the first term is always employed once as a factor, or, in other words, any term of a geometrical series is the product of the ratio, raised to a power whose index. is one less than the number of the term, multiplied by the first term.

NOTE. If the 2d power of a number, as 22, be multiplied by the 3d power, 23, the product is 25. Thus 22 -=2×2=4, and 23=2×2×2=8, and 8x4=32=2×2. ×2×2×2, and, generally, the power produced by multiplying one power by another is denoted by the sum of the indices of the given powers. Hence, in finding the higher powers of numbers, we may abridge the operation by employing as factors several of the lower powers, whose indices added together will make the index of the required power. To find the 7th power of 2, we may multiply the 3d and 4th powers together, thus: 27 = 28: X 24 =8x16=128 Ans.

I. The first term and ratio given to find any other term.

RULE.--Find the power of the ratio, whose index is one less than the number of the required term, and multiply this power by the first term, the product will be the answer, if the series is increasing, but if it is decreasing, divide the first term by the power.

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282. A person bought 6 brooms, giving 3 cents for the first and 96 cents for the last, and the prices form a geometrical series, the ratio of which was 3; what was the cost of all the brooms?

The price would be the sum of the following series: 3+6+12+24+ 4896 189 cents, Ans. If the foregoing series be multiplied by the ratio, 2, the product is, 6+12+24+48+96=192 whose sum is twice that of the first. Now subtracting the first series from this, the remainder is 192 -3-189-the sum of the first series. Had the ratio been other than 2, the remainder would have been as many times the sum of the series as the ratio, less 1, and the renrainder is always the difference between the first term and the product of the last term by the ratio. Hence,

II. The first and last term and ratio given to find the sum of

the series.

RULE.-Multiply the last term by the ratio, and from the product subtract the first term, the remainder divided by the ratio, less 1, will give the sum of the series.

2. The first term of a geo- | second, $4 the third, and so on, metrical series is 4, the last each succeeding payment beterm 972, and the ratio 3;ing double the last; and what what is the sum of the series ?

3-1)9723-4(=1456 Ans.

NOTE. The marks drawn over the numbers show that 4 must be taken from the product of 972 by 3, and the remainder divided by (3-1 =) 2. This mark is called a vincu tum.

3. The extremes of a geometrical progression are 1024 and 59049, and the ratio 1; what is the sum of the series? Ans. 175099.

4. What debt will be discharged in 12 months by paying $1 the first month, $2 the

will be the last payment?

S$4095 the debt. Ans. {$2048 last pay't.

5. A gentleman being asked to dispose of a horse, said he would sell him on condition of having 1 cent for the first nail in his shoes, 2 cents for the second, 4 cents for the third, and so on, doubling the price of every nail to 32, the number of nails in his four shoes; what was the price of the horse at that rate? Ans. $42949672.95.

283. If a pension of $100 dollars per annum be forborn 6 years, what is there due at the end of that time, allowing compound interest at 6 per cent. ?

Whatever the time, it is obvious that the last year's pension will draw no interest, it is therefore only $100; the last but one will draw interest one year, amounting to $106; the last but two, interest (compound) for 2 years, amounting to $112,36; and so on, forming a geometrical progression, whose first term is 100, the ratio 1.06, and the sum of this series will be the amount due. To find the last term (281) say, 1.065 × 100 133.82255776, the sixth term; and to find the sum of the series (282) say, 133.82255777 1.06-100-41.8519112256, which divided by 1.06—1—0 06, gives $667,5318576 Ans, or sum due.

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284. A sum of money payable every year for a number of years is called annuity. When the payment of an annuity is forborn, it is said to

be in arrears.

1. What is the amount of an annuity of $40, to continue 5 years, allowing 5 per cent. compound interest ? Ans. $221.025.

2. If a yearly rent of $50 be forborn 7 years, to what does it amount at 4 per cent. compound interest? Ans. $394.91.

3. Duodecimals.

285. Of the various subdivisions of a foot, the following is one of the most common.

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forming a decreasing geometrical progression, whose first term is 1, and ratio 12. Hence they are called Duodecimals.

286. How many square feet in a floor, 10ft. 4in. long, and 7ft. 8in. wide?

10ft. 4'

7 8

6 10 8 72 4

the

Here we wish to multiply 10ft. 4' by 7ft. 8'; we therefore write them as at the left hand, and multiply 4 by 8-32, but 4' being of a foot, and 8' T product is (X) of a foot, or 32" which reduced gives 2' 8"; putting down 8" we reserve the 2' to be 79ft. 2'8" Ans. added to the inches. Multiplying 10ft. by 8' the product is (223) fg, to which being added we have 6ft. 10. Next, multiplying 4' by 7-3-2ft. 4', writing the 4' in the place of inches, and reserving the 2ft., we say 7 times 10 are 70, and two added are 72, which we write under the 6ft. and the sum of these partial products is 79ft. 2′ 8′′′ Ans.

1

NOTE. When feet are concerned, the product is of the same denomi nation as the term multiplying the feet; and when feet are not concerned, the name of the product will be denoted by the sum of the indices of the two factors, or strokes over them. Thus 4′28′′'. Therefore,

287. To multiply a number consisting of feet, inches, seconds, &c. by another of the same kind.

RULE. Write the several terms of the multiplier under the corresponding terms of the multiplicand; then multiply the whole multiplicand by the several terms of the multiplier successively, beginning at the right hand, and placing the first term of each of the partial products under its respective multiplier, remembering to carry one for every 12 from a lower to the next higher denomination, and the sum of these partial products will be the answer, the left hand term being feet, and those towards the right primes, seconds, &c.

This is a very useful rule in measuring wood, boards, &c. and for artificers in finding the contents of their work.

QUESTIONS FOR PRACTICE.

2. How much wood in a load 7ft. 6′ long, 4ft. 8′ wide, and 4ft. high?

Ans. 140ft. or 1 cord 12ft.

Multiply the length by the width, and this product by the height.

3. How many square feet in a board 16ft. 4in. long, and 2ft. 8in. wide? Ans. 43ft. 6in. 8". 4. How many feet in a stock of 12 boards 14ft. 6' long, and 1ft 3' wide? Ans. 217ft. 6'.

NOTE.-Inches, it will be recollected, are so many 12ths of a foot. whether the foot is lineal, square, or solid. 6in. in the above answer is a square foot, or 72 square inchee.

5. What is the content of a oeiling 43ft. 3' long, and 25ft. 6' broad? Ans. 1102ft. 10′ 6′′.

6. How much wood in a load 6ft. 7' long, 3ft. 5' high, and 3ft. 8' wide?

Ans. 82ft. 5' 8" 4"".

7. What is the solid content of a wall 53ft. 6' long, 12ft. 3' high, and 2ft. thick?

Ans. 1310ft. 9. 8. How many cords in a

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9. How many square yards in the wainscoting of a room 18ft. long, 16ft. 6′ wide, and 9ft. 10 high?

Ans. 75yd. 3ft. 6'.

10. How much wood in a cubic pile measuring 8ft. on every side? Ans. 4 cords.

11. How many square feet in a platform which is 37 feet 11 inches long, and 23 feet 9 inches broad?

Ans. 900ft. 6' 3".

12. How much wood in a load 8 feet 4 inches long, 3ft. 9in. wide, and 4ft. 5in. high? Ans. 138ft. 0' 3".

13. How many feet of flooring in a room which is 28 feet 6 inches long, and 23 feet 5in. broad? Ans. 667ft. 4' 6".

14. How many square feet are there in a board which is 15 feet 10 inches long, and 94 inches wide?

Ans. 12ft. 10′ 4′′ 6".

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