such multiplications, or the product of the first term, 3, by 27, or 3 X 128= 384.

If the last term had been given and the first required, the process would evidently have been by division, since every less term is the result of a division of the term next larger by the ratio.

RULE. Raise the ratio to the power whose index is one less than the number of terms; by which multiply the least term to find the greatest, or divide the greatest to find the least.

NOTE 1.

When the ratio requires to be raised to a high power, the process may be abridged, as in Art 516.

NOTE 2.

- The rule may be applied in computing compound interest, the principal being the first term, the amount of one dollar for one year the ratio, the time, in years, one less than the number of terms, and the amount the last

term.

EXAMPLES.

2. If the first term be 5, and the ratio 3, what is the seventh term? Ans. 3645. 3. If the series be 72, 24, 8, &c., and the number of terms 6, what is the last term?

Ans.

4. If the larger extreme be 885735, the ratio and the number of terms 12, what are the tenth and the eleventh terms? Ans. 45 and 15.

5. If the seventh term is 5, and the ratio, what is the first term? Ans. 3645. 6. If the first term is 50, the ratio 1.06, and the number of terms 5, what is the last term? Ans. 63.123848. 7. If I were to buy 30 oxen, giving 2 cents for the first ox, 4 cents for the second, 8 cents for the third, &c., what would be the price of the last ox?

Ans. $10737418.24.

8. What is the amount of $160.00 at compound interest for Ans. $226.96305796096.

6 years? 9. What is the amount of $300.00 at compound interest at 5 per cent. for 8 years? Ans. $ 443.236+.

10. What is the amount of $ 100.00 at compound interest at 6 per cent. for 30 years?

Ans. $574.349117291325011626410633231080264584635

7252196069357387776.

566. To find the sum of a series, the first term, the ratio, and the number of terms being given.

Ex. 1. If the first term be 1, the ratio 3, and the number Ans. 121.

of terms 5, required the sum of the terms.

OPERATION.

(81 × 3) 3 1

1

= 121.

If we multiply the series 1, 3, 9, 27, 81 by the ratio 3, we shall obtain as a second series 3, 9, 27, 81, 243, whose sum is three times the sum of the first series, and the difference between whose sum and the sum of the first series is evidently twice the sum of the first series. Now it will be observed that the two series have their terms alike, with the excep tion of the first term in the first series, and the last in the second series. We have then only to subtract the first term in the first series from the last term in the second, and the remainder is twice the sum of the first series; and half this being taken gives the required sum of the series. Therefore the sum of the first series must be 242 = 121.

RULE.

2.

Find the last term as in Art. 565, multiply it by the ratio, and the product less the first term divide by the ratio less 1; the result will be the sum of the series.

NOTE 1. - If the ratio is less than a unit, the product of the last term mulplied by the ratio must be subtracted from the first term, and the remainder divided by unity or 1 decreased by the ratio.

NOTE 2. When a descending series is continued to infinity, it becomes what is called an infinite series, whose last term must always be regarded as 0, and its ratio as a fraction. To find the sum of an infinite series,

Divide the first term by 1 decreased by the fraction denoting the ratio, and the quotient will be the sum required.

This process furnishes an expeditious way of finding the value of circulating decimals, since they are composed of numbers in geometrical progression, whose common ratios are 10, 100, 1000, &c. according to the number of factors contained in the repetend. Thus, .3333, &c. represents the geometrical series 10 100 1000, &c. whose first term is and common ratio

3

3

EXAMPLES.

2. The first term of a series is 5, the ratio, and the number

of terms 6; required the sum of the series.

BAPA (1-3)= = 13169, Ans.
66

3325 ÷
729

3. Find the value of the

23

3325
243

180) 100 ÷ (1

4. What is the sum of the series 4, 1, 1, 1, &c., continued to an infinite number of terms?

5. If the first term is 50, the ratio 1.06, and the number of

terms 4, what is the sum of the series?

Ans. 218.7303.

6. A gentleman offered a house for sale on the following terms; that for the first door he should charge 10 cents, for the second 20 cents, for the third 40 cents, and so on in a geometrical ratio. If there were 40 doors, what was the price of the house? Ans. $109951162777.50. 7. If the series 3, 3, 6, 1, 24, &c. were carried to infinity, what would be its sum? Ans. 14.

1

8. A gentleman deposited annually $ 10 in a bank, from the ime his son was born until he was 20 years old. Required the amount of the 21 deposits at 6 per cent., compound interest, when his son was 21 years old? Ans. $423.92.

9768.

9. Find the value of .008497133, &c., continued to infinity. Ans. 83 10. If a body be put in motion by a force which moves it 10 miles in the first portion of time, 9 miles in the second equal portion, and so, in the ratio of 1, for ever, how many miles will it pass over? Ans. 100 miles.

9

567. To find the ratio, the extremes and number of terms being given.

Ex. 1. If the extremes of a series are 3 and 192, and the number of terms 7, what is the ratio?

1923

OPERATION.

64; 64 =

It has been shown (Art. 2, Ans. 565) that the last term of a geometrical series is equal to

the product of the ratio by the first term raised to a power whose index is one less than the number of terms; hence the ratio must equal the root of the quotient of the last term by the first whose index is one less than the number of terms.

RULE. Divide the last extreme by the first, and extract that root of the quotient whose index is one less than the number of terms.

NOTE. When the sum of the series and the extremes are given, the ratio may be found by dividing the sum of all the terms except the first, by the sum of all the terms except the last.

EXAMPLES.

2. If the last term of a series is 1, the largest term 512, and the number of terms 10; what is the ratio?

Ans. 2.

3. If the extremes are 5 and 885735, and the sum of the series 1328600, what is the ratio?

Ans. 3.

4. What debt can be discharged in a year by monthly pay

ments, in geometrical progression, of which the first payment is $ 1, and the last $2048; and what will be the ratio of the series? Ans. Ratio, 2; debt, $ 4095.

568. To insert any number of geometrical means, or mean proportionals, between two given numbers.

Ex. 1. Insert three geometrical means between 4 and 324. Ans. 12, 36, and 108.

terms, the two

Then, having the number

extremes, the number of terms will be 5. of terms and the extremes, we find the ratio, as in the last article, to be 3; and by multiplying the first term by the ratio, we obtain the first of the terms to be inserted. That term multiplied by the ratio gives the next, and that multiplied by the ratio gives the other required mean.

RULE.— Take the two given numbers as the extremes of a geometrical series, and consider the number of terms in the series greater by two than the required number of means. Then find the ratio, as in Art. 567, and the product of the ratio and the first extreme will give one of the means, and the product of this mean and the ratio will give another, and

so on.

NOTE.

When only a single mean is required to be inserted, it may be found as in Art. 550; when only two, as in Art. 551.

EXAMPLES.

2. Insert three geometrical means between

and 128.

Ans. 2, 8, and 32.

3. Required five mean proportionals between the numbers 3 and 2187. Ans. 9, 27, 81, 243, and 729.

569. To find the number of terms, the extremes and ratio being given.

Ex. 1. If the extremes are 5 and 3645, and the ratio 3, what is the number of terms?

OPERATION.

3645729; 36

6 +1 7, Ans.

729.

By Art. 565 it is seen that the ratio raised to the power whose index is one less than the number of terms, and multiplied by the least term, equals the

largest term; hence, the largest term divided by the least term will equal a power of the ratio whose index is one less than the number of terms.

RULE.

Divide the largest term by the least; involve the ratio to a power equal to the quotient; and the index of that power, increased by 1, will be the number of terms.

EXAMPLES.

2. If the extremes are 5 and 20480, and the ratio 4, what is the number of terms? Ans. 7.

3. In what time will a certain debt be discharged by monthly payments in geometrical progression, if the first and last payments are $1 and $ 2048, and the ratio 2?

Ans. In 12 months.

ANNUITIES.

570. ANNUITIES are fixed sums of money payable at the ends of equal periods of time, such as years, or half-years. Annuities in perpetuity are such as continue for ever.

Annuities certain are such as commence at a fixed time, and continue for a certain number of years.

Annuities contingent are those whose commencement or continuance, or both, depend on some contingent event, as the death of one or more individuals.

Annuities deferred, or in reversion, are such as do not commence till after a fixed number of years, or till after some particular event has taken place.

571. An annuity forborne, or in arrears, is one whose periodical payments, instead of being paid when due, have been allowed to accumulate.

572. The amount of an annuity at compound interest, at any time, is the sum to which it will amount, supposing it to have been improved at compound interest during the intervening period.

573. The present value of an annuity at compound interest, for any given period, is the sum of the present values of all the payments of that annuity.