which the ratio ($1.06), the number of terms (5), and the greater term are given, to find the less, as in 183.

TABLE,

Showing the present worth of $1, or £1, from 1 year to 40, allowing compound discount, at 5 and 6 per cent.

186. The extremes and the ratio given, to find the sum of the series.

1. A man bought 4 yards of cloth, giving 2 cents for the first yard, 6 for the second, 18 for the third, and 54 for the fourth; what does he pay for all?

SOLUTION. We may add together the prices of the several yards thus: 2+6+18+54 = 80.

But in a lengthy series this process would be tedious; we will therefore seek for a shorter method. Writing down the terms of the series, we multiply the first term by the ratio, and place the product over the second, to which it will be equal, since the

second term is the product of the first into the ratio. Multiply, also, the second term, placing the product over its equal, the third; multiply the third, placing the product over the fourth; multiply the fourth, and place the product at the right of the last product, thus:

The second series is three times the first series, and subtracting the first from it, there will remain twice the first series. But the terms balance each other, except the first term of the first series, the sum of which we wish to find, and the last term of the second, which is 3 times the last term of the series whose sum we wish. Subtracting the former from the latter, we have left 160, twice the sum of the first, which dividing by 2, the quotient is 80, sum of the series required. Hence,

RULE.

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

EXAMPLES.

2. If the extremes be 4 and 131072, and the ratio 8, what is the sum of the series? Ans. 149796.

3. What is the sum of the decreasing series, 3, 1, 3, 3, 27, &c., extended to infinity?

Such a series is called an infinite series, the last term of which is so near nothing that we regard it 0; hence, when the extremes are 3 and 0, and the ratio 3, what is the sum of the series? Ans. 4.

4. What is the value of the infinite series, 1+1, +16+64

&c.?

Ans. 1.

5. What is the value of the infinite series, +ro, &c., or, what is the same, the decimal continually repeated?

11111, &c., Ans..

6. What is the value of the infinite series, To, Toooo, &c., decreasing by the ratio 100, or, which is the same, the repeating decimal 020202, &c.? Ans..

187. The first term, ratio, and number of terms given, to find the sum of the series.

1. A lady bought 6 yards of silk, agreeing to pay 5 cents for the first yard, 15 for the second, and so on, increasing in a threefold proportion; what did the whole cost?

SOLUTION. We may find the prices of the several yards, and add them together, or, having found the last term by 183, we can find the sum by the last paragraph. But our object is to find a still more expeditious method. Let us find the several terms and write them down as a first series, and below it write a series which we will call the second, having 1 for the first term, and the same number of terms, thus:

First series, 5 15 45 135 405 1215

Third series,

3 9 27 81

243

6th power of ratio.

729

Second series, 1 3 9 27 81 243 729-1=728, which÷2=364, and 364 × 5=1820 cents.

Now multiplying the second series by the ratio, 3, and writing the products as directed in the last paragraph, we have a series three times the second. The last term of the third series, it must be carefully noticed, is the 6th power of 3, the ratio, the power denoted by the number of terms. Subtracting the second series from the third, which is done by taking 1 from the last term of the third, the other terms balancing, 729-1=728, we have twice the second series; and dividing 728 by 2, 728÷2= 364, we have once the second series. Now the first series, the sum of which is required, is 5 times the second, since, as the first term is 5 times greater, each term is 5 times greater than the corresponding term of the second series; and multiplying 364, the sum of the second, by 5, we have the required sum, or 1820 cents = Ans. $18.20.

Hence, The first term, ratio, and number of terms being given, to find the sum of the series,

RULE.

Raise the ratio to a power whose index is equal to the number of terms, from which subtract 1, and divide the remainder by the ratio less 1; the quotient is the sum of a series with 1 for the first term; then multiply this quotient by the first term of any required series; the product will be its amount.

EXAMPLES.

2. A gentleman, whose daughter was married on a Newyear's day, gave her a dollar, promising to triple it on the first day of each month in the year; to how much did her portion amount?

Applying this rule to the example,

312531441, and

531441-1
3-1

x1265720.

Ans. $265720.

3. A man agrees to serve a farmer 40 years without any other reward than 1 kernel of corn for the first year, 10 for the second year, and so on, in tenfold ratio, till the end of the time; what will be the amount of his wages, allowing 1000 kernels to a pint, and supposing he sells his corn at 50 cents per bushel? 1040-1 1,111,111,111,111,111,111,111,111, 111,111,111,111,111 kernels.

10-1

x1 =

{

Ans. $8,680,555,555,555,555,555,555,555,555,555,555-555-5.

4. A gentleman, dying, left his estate to his five sons; to the youngest, $1000, to the second, $1500; and ordered that each son should exceed the younger by the ratio of 11⁄2: what was the amount of the estate?

NOTE 1.-Before finding the power of the ratio 11, it may be reduced to an improper fraction = 3, or to a decimal, 1·5.

35-1

×1000-$131871; or,

1.55-1
1.5-1

× 1000-$13187.50.

Ans. $13187.50.

Questions.-185. Ex. 1, and its solution? Note 1? 186. Ex. 1. What are given to find what? Solution on the Blackboard? Rule? Ex. 8, is an example of what? Infinite series? 187. What are given to find what? Ex. 1, and its solution? Rule?

188. Annuities at Compound Interest.

1. A man rented a dwelling-house for $100 a year, but did not receive any thing till the end of 4 years, when the whole was paid, with compound interest at 6 per cent. on the sums not paid when due; what did he receive?

SOLUTION. AS annuities in arrears at simple interest form an arithmetical series, so the several years' rent, with compound interest on those in arrears, are so many terms of a geometrical series. The last year's rent is $100 only, since it is paid when due, at the end of the year; the third year's rent is on interest 1 year, and is found by multiplying $100 by 1.06, producing $106, and this product multiplied by 106, will give the second year's rent, paid 2 years after it is due, and so on. The first term, $100, the number of terms, 4, and the ratio, 106, are given to find the sum, as in the last paragraph, and we may apply the same rule, thus:

1.06'- 1
⚫06

X 100437.45.

Ans. $437.45.

NOTE 1.-The powers of the ratio, see 184, may be found in the table (132).

EXAMPLES.

2. What is the amount of an annuity of $50, it being in arrears 20 years, allowing 5 per cent. compound interest? Ans. $1653.29.

3. If the annual rent of a house, which is $150, be in arrears 4 years, what is the amount, allowing 10 per cent. compound interest? Ans. $696.15.

4. To how much would a salary of $500 per annum amount in 14 years, the money being improved at 6 per cent. compound interest? in 10 years? in 20 years? years?

in 24 years?

in 22

Ans. to the last, $25407.75.

5. Two men commence life together; the one pays cash down,

$200 a year to mechanics and merchants; the second gets precisely the same value of articles, but on credit, and proving a negligent paymaster, is charged 20 per cent. more than the other; what is the difference in 40 years, compound interest being calculated at 6 per cent.? Ans. $6190 478+.