RULE.

Multiply half the sum of the extremes by the number of terms. The product will be the sum of the series.

14. A man has 9 sons; the youngest is 8, the eldest 40 What is the sum of their ages? years old. Ans. 216 years.

15. How many times will a clock strike in a day, if constructed like the clocks of Venice, to run till 24 o'clock ?

Ans. 300. 16. If a triangular piece of land, 60 rods in length, be 1 rod wide at one end, and 60 at the other, what number of square rods does it contain? Ans. 1830.

GEOMETRICAL PROGRESSION.

Art. 241.-A GEOMETRICAL PROGRESSION is a series of terms, which increase by a uniform multiplier, or decrease by a uniform divisor; as 3, 6, 12, 24, etc., increasing by a uniform multiplier, 2; or 54, 18, 6, 2, 3, etc., decreasing by a uniform divisor, 3.

The multiplier, or divisor, which produces the series, is called the ratio.

1. A man bought 5 yards of cloth, paying 3 cents for the first yard, 6 for the second, and so on, doubling the price to the last. What was the price of the last yard?

3×2×2×2×2=48, the cost of the last yard.

From the above operation it will be seen that the cost of the second yard is the product of the ratio multiplied by the cost of the first yard; and that the cost of the third yard, is the product of the second power of the ratio multiplied by the cost of the first yard, or the first term; and finally, that the cost of the fifth yard, or the last term, is the product of the fourth power of the ratio, multiplied by the cost of the first yard. It appears, also, that any term in the series may be found by in

QUESTIONS.-1. What is Geometrical Progression? 2. What is an ascending series? 3. What a descending? 4. What is the ratio? 5. When the first term and ratio are given, how do you find the last term? 6. When the first and last terms, and the ratio are given, how do you find the sum of the series?

volving the ratio to a power less 1 than the number of terms, and multiplying that power by the first term.

OBS.-The process of involving the ratio to a high power, may be shortened by multiplying together those lower powers whose indices added equal the index of the power sought. To find the fifth power of 3, we may multiply together the second and third powers, for the index of the second power of 3, and the index of the third power added,

equal 5, 3 3 and 33X33-9 X 27-243, the 5th power of 3. Art. 242.-When the first term and ratio are given, to find the last term

RULE.

Involve the ratio to a power whose index is 1 less than the number of terms, 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 ratio.

2. If I hire a man for 12 months, and agree to pay him 1 dollar for the first month, 3 for the second, and so in a triple proportion, what must I pay him for the last month?

1 2 3 4

6

6

5 =

11

3 3 27 81 243 729, 729243=177147×1=$177147 Ans. It will be seen that the sum of the indices of the fifth and sixth powers added, equal 11, which is 1 less than the number of terms; and the fifth and sixth powers multiplied together, equal the 11th power of the ratio, which multiplied by the first term, gives the answer, or the last term.

3. A man bought 20 cows, paying 2 farthings for the first, 10 for the second, and so on, in a five-fold ratio. What was the price of the last cow?

Ans. £39736429850 5s. 2d. 2qrs.

4. A man bought 5 yards of cloth, giving 2 cents for the first, and 32 for the last; the prices forming a geometrical series, the ratio of which was 2. What was the whole cost of the cloth?

The price of the cloth would be the sum of the following numbers: 2+4+8+16+32=62, the whole cost. It will be seen, that the whole cost is the same as the difference between the two extremes divided by the ratio less 1 added to the greater extreme: Thus, 32-2 = 30, and 301 = 30, and 30+32=62.

We

Again, if any term of a corresponding series be multiplied by the ratio, the product will be the succeeding term. will now form a new series, and write it one step farther to the right of that from which it is formed; if we now subtract the first series from the second, we find that all the terms but the first in the first series and the last in the second, disappear, thus:

OBS.-If the ratio were 3 we should have double the first series, if it were 4 we should have triple; hence we divide by the ratio less one.

Art. 243. Hence, when the first term, last term, and ratio, are given, to find the sum of the series, we have the following

RULES.

I. Multiply the last term by the ratio, and from the product subtract the first term, and divide the remainder by the ratio less 1; the quotient will be the answer.

II. Divide the difference between the two extremes by the ratio less 1, and add the quotient to the greater term; their sum will be the answer.

5. The extremes of a Geometrical Progression are 3 and 18673, and the ratio 11. What is the sum of the series?

6. If I discharge a debt by paying 1 dollar the first month, 4 the second, and so on, in a four-fold ratio, the last payment being 65536 dollars; what was the whole debt?

Ans. $87381.

7. The first term in a geometrical series is 2, the number of terms 10, and the ratio 3. What is the sum of the series?

Ans. 59048.

OBS. The last term may be found by rule first, or the two processes of finding the last term and the sum of the series may be reduced to one,

thus:

Art. 244.-When the first term, the number of terms, and the ratio are given, to find the sum of the series—

RULE.

Involve the ratio to a power whose index is equal to the number of terms, from which subtract 1; divide the remainder by the ratio less 1, and the quotient, multiplied by the first term, will

be the answer.

8. A man sold 15 yards of cloth; the first yard for 1 shilling, the second for 2, the third for 4, and so on, doubling the price of each succeeding yard. For how much did he sell the whole?

Operation.

215-32768, and 32768—1×1=32767s.

2-1

20)3276|7
£1638 78. Ans.

9. A man bought 20 yards of cloth, agreeing to pay 3 pence for the first yard, 9 pence for the second, and so on in a triple proportion to the last. What did he pay for the whole ? Ans. £21792402 10s. 10. A gentleman bought a horse, agreeing to pay what his shoes would amount to, at 1 cent for the first nail, 2 for the second, 4 for the third, and so on, doubling the price of each succeeding nail to the last. The number of nails was 32; what was the price of the horse? Ans. $42949672.95. 11. A laborer wrought 20 days, and received for the first day's labor 4 grains of rye, for the second 12, for the third 36, &c. How much did his wages amount to, allowing 7680 grains to make a pint, and the whole to be disposed of at $1 per bushel? Ans. $14187+.

COMPOUND INTEREST BY PROGRESSION.

Art. 245.-1. What is the amount of $6, for four years, at 6 per cent., compound interest ?

It will be seen, that this question may be solved by the rule after Example 1st in Progression. The principal is the first term, the amount of $1 for one year the ratio, and the number of years, 1 less than the number of terms. The question may be thus stated:-If the first term be 6, the number of terms 5, and the ratio 1.06, what is the last term?

1.06-1.262 × 6-7.572 dollars.

The amount of £1, or $1, at 5 or 6 per cent., may be found by the table for Compound Interest, (see Art. 211.)

2. What is the amount of $30, for 7 years, at 5 per cent., compound interest ? Ans. $42.213. 3. What is the amount of $7, for 4 years, at 9 per cent., compound interest? Ans. $9.881.

4. If the amount of a certain sum for 6 years, at 6 per cent., compound interest, be $56.74040, what is that sum, or principal?

It will be seen that this question is the reverse of the preceding. If the amount be the product of that power of the ratio denoted by the number of years and the principal, then the amount divided by that power of the ratio will be the principal.

56.74040

1.066

$40 Ans.