ference and number of terms given, how do you find the last term?

9. Having the first term, last term, and the number of terms given, how do you find the common difference?

10. Having the first term, last term,

and number of terms given, how do you find the sum of all the terms?

11. Having the first term, last term and common difference, how do you find the number of terms?

GEOMETRICAL PROGRESSION,

Is any rank or series of numbers, increasing by a constant multiplier, or decreasing by a constant divisor; and this multiplier or divisor is called the ratio of the progression.

Thus,

(1, 2, 4, 8, 16, &c., is an increasing geometrical series; and

8, 4, 2, 1,

&c., is a decreasing geometrical series, and the ratio is 2.

In this, as in arithmetical progression, there are five parts, viz.: 1st, the first term; 2nd, the last term; 3d, the number of terms; 4th, the ratio; 5th, the sum of all the terms, any 3 of which being given, the other two may be found.

CASE 1.

Given, the first term, the ratio, and the number of terms, to find the last term.

RULE..

1. Write down a few of the leading powers of the ratio, placing their indices,* viz.: 1, 2, 3, 4, &c. over them.

2. Add together the most convenient indices, to make an index, less by 1 than the number of the term sought.†

3. Multiply together the powers belonging to, or standing under, those indices, and their product, multiplied by the first term, will give the term sought.

* To find the last term of a long series of numbers, by multiplication, would be very tedious, therefore we have a series of numbers in arithmetical proportion, called indices, whose common difference is 1. Then if the powers standing under any of those indices be multiplied together, their product will express a power equal to the sum of those indices which were used.

+ The reason why the sum of the indices must be 1 less than the number of terms sought is very evident: for example, if the first term be 4, and the ratio 3, required the sixth term. Thus, 4×3×3×3×3×3=972, the sixth term; and if we examine the process, we shall see that the ratio is five times a factor; that is, one time less than the number of terms; and the fifth power of the ratio, mul

EXAMPLES.

1. A man bought 12 yards of clotn, and by agreement was to pay what the last yard would come to, at 3 cents for the 1st yard, 6 cents for the second, and so on, doubling the price to the last; what did the last yard come to?

2 4 8 16 32 leading powers of the ratio. Then, 5+4+2=11, the number of terms, less 1. 32×16×4=2048, the 11th power of the ratio. X3, the first term.

The last term, or

cost of the last yard,} 6144cts.—$61,44, Ans.

2. İf the first term be 4, and the ratio 3, what is the

25th term?

Ans. 1129718145924.

3. A drover purchased 15 head of cattle, and agreed to pay what the last would come to, reckoning 3 dollars for the first, 12 dollars for the second, and so on in a quadruple, or fourfould proportion; I demand the sum to be paid? Ans. $805306368.

4. A draper sold 20 yards of superfine cloth, on condition that he should receive pay for the last yard only, reckoning 3 cents for the first yard, 9 cents for the second, 27 cents for the third, and so on in triple proportion; how much did he receive for the whole?

Ans. $34867844,01.

The first had

5. Twelve men received a sum of money. $3, the second $6, the third $12, and so on in a twofold proportion; how many dollars had the last? Ans. $6144.

6. If the first term of a geometrical series be 1, and the ratio 6, what is the 12th term? Ans. 362797056.

tiplied by the first term, produces the sixth. Hence the sum of the indices used denoting the leading powers of the ratio, must always be 1 less than the number of the terms sought.

Again; if the first term be 3, and the ratio 2, what is the 13th term 2-Thus, 3×2×2×2×2×2×2×2×2×2×2×2×2=12288. Here we see that the ratio is twelve times a factor, and multiplied into the first term; that is, the 12th power of the ratio multiplied by the first term, -13th term.-But to raise the ratio to its 12th power, we need not multiply all the intermediate powers; for 24-16 and 16 x 16-256, the 8th power; that is, the 4th power x 4th power- 8th power, and 256 × 16-4096; the 12th power, and 4096x3=12288, as before.

CASE 11.

The first term, the last term, (or the extremes,) and the ratio given, to find the sum of the series.

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

EXAMPLES.

1. A man bought 6 yards of cloth, giving 2 cents for the first yard, 6 cents for the second, and so on in triple proportion; what did the last yard cost, and how much was the cost of the whole ?

By Case I, the cost of the last yard is 486 cents.

2. If the first term of a geometricial series be 3, and the last term 6144, and the ratio 2, what is the sum of all the terms? Ans. 12285. 3. If the extremes of a geometrical series be 10, and 196830, and the ratio 3, what is the sum of all the terms? Ans. 295240.

(Note. In the following examples, the scholar must find the last term of the series by Case I; then find the sum of all the terms, as above.)

4. A man purchased a valuable tract of land containing fifteen acres, agreeing to give 1 dollar for the first acre, 4 dollars for the second, and so on in a quadruple, or fourfold proportion; what did the whole tract cost him?

Ans. $357913941. 5. What debt can be discharged in a year, by paying 1 cent the first month, 10 cents the second, and so on in a tenfold proportion? Ans. $1111111111,11.

6. If a pound (12oz.) of gold be sold at the rate of 2 cents for the first ounce, 8 cents for the second, 32 cents for the third, and so on in a fourfold proportion to the last; what will it amount to? Ans. $111848,10.

7. A merchant sold 14 yards of Italian silk at the rate of 4 cents for the first yard, 12 cents for the second, and so

on in geometrical progression; how much did the last yard come to, and what did the whole amount to? Ans. The last come to $63772,92, and the whole $95659,36.

8. A man bought a horse, and by agreement was to give a cent for the first nail, two for the second, four for the third, and so on; there were four shoes, and eight nails in each shoe; what did the horse come to at that rate?

Ans. $42949672,95.

9. A thresher worked 20 days for a farmer, and received for the first day's work 4 barley-corns, for the second 12 barley-corns, for the third 36 barley-corns, and so on in triple geometrical progression; what did his 20 day's labour come to, allowing 7680 barley-corns to make a pint, and the whole quantity to be sold at 50cts. per bushel?

Ans. $7093,50, rejecting remainders. 10. If a body put in motion move 1 inch the first second of time, 3 inches the second, 9 inches the 3d second of time, and thus continue to increase its motion in triple proportion, geometrical, how many yards will it move in the term of half a minute? Ans. 2859599056870yds. Oft. 4 inches, which is no less than one thousand six hundred and twenty-four millions of miles.

Questions.

1. What is Geometrical Progression? 2. What is the multiplier or divisor in Geometrical Progression called? 3. How many parts are there in Geometrical Progression?

4. Having the first term, the ratio,

and the number of terms given, how do you find the last term?

5. Having the first term, the last term, (or the extremes,) and the ratio given, how do you find the sum of the series?

ANNUITIES.

To find the amount of an annuity at Simple Interest, by Arithmetical Progression.

RULE.

Make 1 the first term of an arithmetical series, and the ratio the common difference-multiply the common difference by the number of terms less 1, and to the product add the first term for the last. Then find the sum of all the terms by the Rule, Case III. page 205, which will be the amount of $1 annuity for the given number of years Multiply this amount by the given sum for the whole amount

EXAMPLES.

1. What will $200 yearly annuity, remaining unpaid or in arrears 8 years, amount to at 6 per cent?

2. What is the amount of a yearly rent of $75 remaining unpaid, or in arrears, 25 years, at 6 per cent? [Ans. ,06×24+1=2,44. Then 1+2,44×25÷÷2 × 75=$3225, 3. What is the amount of an annuity of $600 remaining in arrears 10 years, at 6 per cent ? Ans. $7620.

ANNUITIES OR PENSIONS COMPUTED AT COMPOUND

INTEREST.

CASE I.

To find the amount of an annuity or pension in arrears at Compound Interest.

RULE.

Raise the ratio to a power equal to the given number of years. From that power subtract 1, and divide the remainder by the ratio, less 1, and the quotient will be the amount of $1 annuity for the given time. Multiply this amount by the given annuity, and the product will be the amount required.

Note. The ratio is the amount of $1, &c. at the given rate for one year.

EXAMPLES.

1. What will $60 yearly annuity amount to, being forborne, or unpaid 5 years, at 6 per cent compound interest? Ratio 1,06×1,06 × 1,06×1,06 × 106—1,÷,065,63709+. Amount of $1 annuity for 5 years=5,63709

given annuity

X 60

Amount of $60 annuity 5 years $338,22540