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15. If a piece of land, 60 rods in length, be 20 rods wide at one end, and at the othar terminate in an angle or point, what number of square rods does it contain? A. 600.

16. A person, travelling into the country, went 3 miles the first day, and increased every day's travel 5 miles, till at last he went 58 miles in one day; how many days did he travel ?

We found, in example 1, the difference of the extremes, divided by the number of terms, loss 1, gave the common differenc: : consequently, if, in this example, we dívide (58 - 3=) 55, the difference of the extremes, by the common difference, 5, the quotiedt, 11, will be the number of terms, less 1; then, 1+ ll = 12, the number of terms. A. 12.

Hence, When the Extremes and Common Difference are given,

to find the Number of Terms ;Divide the difference of the extremes by the common difference, and the quotient, increased by 1, will be the

answer.

17. If the extremes be 3 and 45, and the common difference 6, wha: is the number of terms ? A. 8.

18. A man, being asked how many children he had, replied, tha: the youngest was 4 years old, and the eldest 32, the increase of the family having been i in every 4 years ; how many had he ? A. 8.

GEOMETRICAL PROGRESSION. 1 LXXXIX. Any rank or series of numbers, increasing by a constant multiplier, or decreasing by a constant divisor, is called Geometrical Progression.

Thus, 3, 9, 27, 81, &c., is an increasing geometrical series ;

And 81, 27, 9, 3, &c., is a decreasing geometrical series.

There are five terms in Geometrical Progression; and, like Arithmetical Progression, any three of them being given, the other two may be found, viz.

1. The first term.
2. The last term.
3. The number of terms.
4. The sum of all the terms.

5. The ratio ; that is, the multiplier or divisor, by which we form the series.

1. A man purchased a flock of sheep, consisting of 9; and, by agreement, was to pay what the last sheep came to, at the rate of $4 for the first sheep, $12 for the second, $36 for the third, and so on, trebling the price to the last; what did the Apck cost hun?

Wo may perform this example by multiplication ; thus, 4 X3X3X3X3 X3 X3 X3 X 3 = $20244, Ans. But this process, you must bo sensible, would be, in many cases, a very tedious one ; let us see if we cannot abridge it, thereby making it easier.

In tho above process, we discover that 4 is multiplierl by 3 eight times, one time less than the number of terms; consequently, the 8th power of the ratio 3, expressed thus, 38, multiplied by the first term, 4, will produce the last term. But, instrad of raising 3 to the 8th power in this manner, we neeil only raise it to the 4th power, then multiply this 4th power into itself; for, in this way, we do, in fact, use the 3 cight times, raising the 3 to the same power as before ; thus, 34 =81; then, 81 X 81 = 6551; this, multiplied by 4, the first term, gives $26244, the same result as before. A. 26244.

Hence, When the First Term, Ratio, and Number of Terms,

are given, to find the Last Term ;Write down some of the leading powers of the ratio, with the numbers 1, 2, 3, &c., over them, being their several indices.

Add together the most convenient indices to make an index less by 1 than the number of terms sought.

Multiply together the powers, or numbers standing under those indices; and their product, multiplied by the first term, will be the term sought.

2. If the first term of a geometrical series be 4, and the ratio 3, what is the 11th term ?

1, 2, 3, 4, 5, indices. Note. The pupil will notice that the series

3, 9, 21, 81, 243, powers. | does not commence with the first term, but with the ratio.

The indices 5+3+2= 10, and the powers under each, 243 X 27 X 9= 59049; which, multiplied by the first term, 4, makes 236196, the 11th term, réquired. A. 236196.

3. The first term of a series, having 10 terms, is 4, and the ratio 3; what is the last term? A. 78732.

4. A sum of money is to be divided among 10 persons; the first to have $10, the second $30, and so on, in threefold proportion; what will the last have?

A. $195830. 5. A boy purchased 18 oranges, on condition that he should pay only the price of the last, reckoning 1 cent for the first, 4 cents for the second, 16 cents for the thiril, and in that proportion for the whole; how much did he pay for them? A. 171798691,84

6. What is the last term of a scries having 18 terms, the first of which is 3, and the ratio 3? A. 387420.189.

7. A butcher meets a drover, who has 24 oxen. The butcher inquires the price of them, and is answered, $60 per head; he immediately offers the drover $50 per heal, and would take all. The drover says he will not take that ; but, if he will give him what the last ox would come to, at 2 cents for the first, 4 cents for the second, and so on, doubling the price to the last, he might have the whole. What will the oxen amount to at that rate?

A. $167772,16. 8. A man was to travel to a certain place in 4 days, and to travel at whatever rate he pleased; the first day he went 2 miles, the second 6 miles, and so oh to the last, in a threefold ratio ; how far did he travel the last day, and how far in all!

In this example, we may find the last term as before, or find it by adding cach day's travel together, commencing with the first, and proceeding to the last, thus: 2.+67 18 + 54 = 80 miles, the whole distance travelled, and the last day's journey is 54 miles. But this mode of operation, in a long series, you must be sensible, would be very troublesome. Let us examine the nature of the series, and try to invent some shorter method of arriving at the same result.

By examining the series 2, 6, 18, 54, we perceive that the last term (54), less 2 (tho first term), = 52, is 2 times as largo as the sum of the remaining terms; for 2+ 6+18=26; that is, 54 – 2 = 52 = 2 = 23; hence, if wo produce another term, that is, multiply 54, the last term, by the ratio 3, making 162, we shall find the same true of this also; for 102 – 2 (the first term), =160=2= 80, which we at first found to be the sum of the four remaining terms, thus, 2 +6+ 18 + 54 = 80. In both of these operations, it is curious to observe, that our divisor (2), each time, is 1 less than the ratio (3).

Hence, When the Extremes and Patio are given, to find the
Sum of the Series, we have the following easy

RULE,
Multiply the last term by the ratio, from the product
subtract the first term, and divide the remainder by the
ratio, less 1; the quotient will be the sum of the series
required.

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9. If the extremes be 5 and 6400, and the ratio 6, what is the whole amount of the series?

6400 X 6-5

=7679, Ans.

6-1 10. A sum of money is to be divided among 10 persons in such a manner, that the first may have $10, the second $30, and so on, in threefold proportion ; what Will the last have, and what will the whole have?

The pupil will recollect how he found the last term of the series by a foregoing rule; and, in all cases in which he is required to find the sum of the series, when the last term is not given, he must first find it by that rule, and then work for the sum of the series by the present rule.

A. The last, $196830 ; and the whole, $295240. 11. A hosier sold 14 pair of stockings, the first at 4 cents, the second at 12 cents, and so on in geometrical progression; what did the last pair bring him, and what did the whole bring him?

A. Last, $63772,92; whole, $95659,36. 12. A man bought a horse, and, by agreement, was to give a cent for the first nail, three for the second, lc.; there were four shoes, and in each shoe eight nails; what did the horse come to at that rate?

A. $9265100944259,20, 13. At the marriage of a larly, one of the guests made her a present of a half-cagle, saying that he would double it on ihe first day of each succeeding month throughout the year, which he said would amount to something liko $100; now, how much did his estimate differ from the true amount :

A. $20375. 14. If our pious ancestors, who landed at Plymouth, A. D. 1620, being 101 in number, had increased so as to double their number in every 20 years, how great would have been their population at the end of the year 1840 ?

A. 206747

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and

ANNUITIES AT SIMPLE INTEREST. q XC. An annuity is a sum of money, payable every year, for a certain number of years, or forever.

When the annuity is not paid at the time it becomes due, it is said to be in arrears.

The sum of all the annuities, such as rents, pensions, &c., remaining unpaid, with the interest on each, for the time it has been due, is called the amount of the annuity.

Hence, To find the Amount of an Annuity ;Calculate the interest on each annuity, for the time it has remained unpaid, and find its amount; then the sum of all these several amounts will be the amount required.

1. If the annual rent of a house, which is $200, remain unpaid (that is, te arrears) 8 years, what is the amount?

In this example, the rent of the last (8th) year being paid when due, ot, course, there is no interest to be calculated on that year's rent.

The amount of $200 for 7 years = $284
The amount of $200 6

$272
The amount of $200

$260 The amount of $200

$248 The amount of $200

$236 The amount of $200

$224 The amount of $200 1

$212 The eighth year, paid when due,= $200

$1936, ARS 2. If a man, having an annual pension of $60, receive no part of it till the expiration of 8 years, what is the amount then due? A. $580,80.

3. What would an annual salary of $600 amount to, which remains unpaid (or in arrears) for 2 years ?-1236. For 3 years ?-1908. For 4 years ?-2016. For 7 years 2-1956. For 8 years ?-5808. For 10 years

2-7620.

Ans. $24144.1 4. What is the present worth of an annuity of $600, 10 continue 4 years?

The present worth (1 LXVII.) is such a sum, as, it was al interest, would amount to the given annuity; hence,

$600 - $1,06 = $566,037, present worto, 1st year.
$600 = $1,12 = $535,714,
$600 = $1,18 = $508,474,
$600 = $1,24 = $483,870,

Ans., $2094,095, present worth required. Hence, To find the Present Worth of an Annuity; Find the present worth of each year by itself, discounting from the time it becomes due, and the sum of all these present worths will be the answer.

5 4 3 2

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tinue 3 years,

5. What sum of roady money is equivalent to an annuity of $200, to con

at 4 per cent.? A. $556,063.
6. What is the present worth of an annual salary of $800, to continue 2
years ?-1469001. 3 gears ?-2146967. 5 years ?-3407512. A. $7623,48.

ANNUITIES AT COMPOUND INTEREST.

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1 XCI. The amount of an annuity, at simple and com-
pound interest, is the same, excepting the difference in
interest
Hence, To find the Amount of an Annuity at Compound

Interest;
Proceed as in 1 XC., reckoning compound, instead of
simple interest.
1. What will a salary of $200 amount to, which has remained unpaid for 3

The amount of $200 for 2 years =$224,72
The amount of $200 for 1 year =

- $212,00
The 3d year..

$200,00

A. $636,72
2. If the annual rent of a house, which is $150, remain in arrears for 3
yours, what will be the amount due for that time? A. 477,54.

Calculating the amount of the annuities in this manner, for a long period of
years, would be tedious. This truulie will be prevented, by finding the
annount of $1, or £1, annuity, at compound interest, for a number of years, as
in the following

TABLE I.

years?

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Showing the amount of $1 or £1 annuity, at 6 per cent. compound in

terest, for any number of years, from 1 to 50.
Yrs. 6 per ct. Yrs.16 per ct.|| Yrs. 6 per ct.) Yrs. 6 per ct. Yrs. 6 por ci.

1 1,0000 il 14,9716 21 39,9927 31 84,8016| 41 165,0467
2
2,0600

12 16,8099 22 43,3922 || 32 90,8897| 42 175,9495|
3 3,1836 13 18,8821

23 46,9958 33 97,3431 43 187,5064
4 4.3746 14 21,0150 || 24 50,8155 34 104,1837|| 44 199,7568
5 5,6371 15 23,2759 | 25 | 54,8645 35 111,4347|| 45 212,7423
6 6,9753 16 25,6725 26 59,1563 36 119,1208|| 46 226,5068
7 8,3938 17 28,2123 27 63,7057 37 (127,2681|| 47 231,0972
8 9,8974 18 30,9056 | 28 | 68,5281 38 135,904248 245,9630

11,4913 1933,7599 29 73,6397 39 145,05841 49 261,7208
10 | 13,1807 20 / 36,7855 30 1 79,0581 40 |151,76191 50 278,4241

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