1

3. The first term is 0, the last term, and the ratio 2; what is the number of terms?

4. The extremes are 196608 and 6, and the ratio is ; what is the number of terms?

Ans. 6.

PROBLEM IV.

714. Given, the extremes and ratio, to find the sum of the series.

Let us take the series +5+20+80+320=425, multiply each term by the ratio 4, and from this result subtract the given series term from term, thus:

20+80+ 320 + 1280 = 1700, four times the series, 5+20+80 + 320 425, once the series,

Then

=

Hence the

RULE. Multiply the greater extreme by the ratio, subtract the less extreme from the product, and divide the remainder by the ratio less 1.

NOTE. Let every descending series be inverted, and the first term called the last; then the ratio will be greater than a unit. If the series be infinite, the least term is a cipher.

EXAMPLES FOR PRACTICE.

1. The extremes are 3 and 384, and the ratio is 2; what is the sum of the series? Ans. 765. 2. If the extremes are 5 and 1080, and the ratio is 6, what is the sum of the series?

1295°

8 4059

3. If the first term is 44, the last term and the ratio, what is the sum of the series? Ans. 75

77

4. What is the sum of the infinite series, 8, 4, 2, 1, 1, 1, etc.?

PROBLEM V.

715. Given, the first term, the ratio, and the num

ber of terms, to find the sum of the series.

If, for example, the first term be 4, the ratio 3, and the number of terms 6, then by Problem I, we have

4 x 35 the last term.

RULE. Raise the ratio to a power indicated by the number of terms, and subtract 1 from the result; then multiply this remainder by the first term, and divide the product by the ratio less 1.

EXAMPLES FOR PRACTICE.

.1. The first term is 7, the ratio 3, and the number of terms 4; what is the sum of the series?

Ans. 280. 2. The first term is 375, the ratio, and the number of terms 4; what is the sum of the series?

468

3. The first term is 175, the ratio 1.06, and the number of terms 5; what is the sum of the series?

986.45+

PROBLEM VI.

716. Given, the extremes and the sum of the series, to find the ratio.

If we take the geometrical progression, 2, 6, 18, 54, 162, in which the ratio is 3, and remove the first term and the last term, successively, and then compare the results, we have

=

6+18+54 + 162 sum of the series minus the first term. 2 + 6 +18+ 54= sum of the series minus the last term, Now, since every term in the first line is 3 times the corresponding term in the second line, the sum of the terms in the first line must be 3 times the sum of the terms in the second line. Hence the

RULE. Divide the sum of the series minus the first term, by the sum of the series minus the last term.

EXAMPLES FOR PRACTICE.

1. The extremes are 2 and 686, and the sum of the series is 800; what is the ratio?

Ans. 7.

2. The extremes are and 64, and the sum of the series is 149; what is the ratio?

3. If the sum of an infinite series be 42, and the greater extreme 3, what is the ratio?

Ans. 3. 717. Every other problem in Geometrical Progression, that admits of an arithmetical solution, may be solved either by reversing or combining some of the problems already given.

COMPOUND INTEREST BY GEOMETRICAL PROGRESSION.

718. We have seen (551) that if any sum at compound interest be multiplied by the amount of $1 for the given interval, the product will be the amount of the given sum or principal at the end of the first interval; and that this amount constitutes a new principal for the second interval, and so on for a third, fourth, or any other interval. Hence,

A question in compound interest constitutes a geometrical progression, whose first term is the principal; the common multiplier or ratio is one plus the rate per cent. for one interval; the number of terms is equal to the number of years plus one; and the last term is the amount of the given principal for the given time. All the usual cases of compound interest and discount computed at compound interest, can therefore be solved by the rules for geometrical progression. For example,

Find the amount of $250 for 4 years, at 6 % compound interest.

OPERATION.

$250 x 1.064 $250 × 1.262477

$316.21925.

ANALYSIS. Here we have $250 the first term, 1.06 the ratio, and 5 the number of terms, to find the last term. Then by 706 we find the last term, which is the amount required.

EXAMPLES FOR PRACTICE.

1. What is the amount of $350 in 4 years, at 6 % per annum compound interest? Ans. $441.86. 2. Of what principal is $150 the compound interest in 2 years, at 7 %?

3. What sum at 6 % compound interest, will amount to $1000 in 3 years? Ans. $839.62.

4. In how many years will $40 amount to $53.24, at 10 % compound interest? Ans. 3 years.

5. At what rate per cent. compound interest will any sum double itself in 8 years?

405+

Ans. 11.04

%.

6. What is the present worth of $322.51, at 5 % compound interest, due 24 years hence? Ans. $100.

ANNUITIES.

719. An Annuity is literally a sum of money which is payable annually. The term is, however, applied to a sum which is payable at any equal intervals, as monthly, quarterly, semi-annually, etc.

NOTE. The term, interval, will be used to denote the time between payments. Annuities are of three kinds: Certain, Contingent, and Perpetual.

720. A Certain Annuity is one whose period of continuance is definite or fixed.

721. A Contingent Annuity is one whose time of commencement, or ending, or both, is uncertain; and hence the period of its continuance is uncertain.

722. A Perpetual Annuity or Perpetuity is one which continues forever.

723. Each of these kinds is subject, in reference to its commencement, to the three following conditions:

1st. It may be deferred, i. e., it is not to be entered upon until after a certain period of time.

2d. It may be reversionary, i. e., it is not to be entered upon until after the death of a certain person, or the occurrence of some certain event.

3d. It may be in possession, i. e., it is to be entered upon at

once.

724. An Annuity in Arrears or Forborne is one on which the payments were not made when due. Interest is to be reckoned on each payment of an annuity in arrears, from its maturity, the same as on any other debt.

ANNUITIES AT SIMPLE INTEREST.

725. In reference to an annuity at simple interest, we observe: I. The first payment becomes due at the end of the first interval, and hence will bear interest until the annuity is settled.

II. The second payment becomes due at the end of the second interval, and hence will bear interest for one interval less than the first payment.

III. The third payment will bear interest for one interval less than the second; and so on to any number of terms. Hence,

IV. All the payments being settled at one time, each will be less than the preceding, by the interest on the annuity for one interval. Therefore, they will constitute a descending arithmetical progression, whose first term is the annuity plus its interest for as many intervals less one as intervene between the commencement and settlement of the annuity; the common difference is the terest on the annuity for one interval; the number of terms is number of intervals between the commencement and settlement of the annuity; and the last term is the annuity itself.

726. The rules in Arithmetical Progression will solve all problems in annuities at simple interest.

EXAMPLES FOR PRACTICE.

1. A man works for a farmer one year and six months, at $20 per month, payable monthly; and these wages remain unpaid until the expiration of the whole term of service. How much is due to the workman, allowing simple interest at 6 per cent. per annum?