Consequently, the greatest term is equal to the least term added to the product of the common difference multiplied by 1 less than the number of terms.

THEOREM 4. The sum of all the terms, of any arithmetical progression, is equal to the sum of the two extremes multiplied by the number of terms, and divided by 2; or the sum of the two extremes multiplied by the number of the terms, gives double the sum of all the terms in the series.

This is made evident by setting the terms of the series in an inverted order, under the same series in a direct order, and adding the corresponding terms together in that order. Thus, in the series 1, 3, 5, 7, 9, ditto inverted 15, 13, 11, 9, 7,

13,

15;

11,
5, 3, 1;

the sums are 16 + 16 + 16 † 16 † 16 ♣ 16 + 16 + 16, which must be double the sum of the single series, and is equal to the sum of the extremes repeated as often as are the number of the terms.

From these theorems may readily be found any one of these five parts; the two extremes, the number of terms, the common difference, and the sum of all the terms, when any three of them are given; as in the following problems:

PROBLEM I.

Given the Extremes, and the Number of Terms; to find the Sum of all the Terms.

ADD the extremes together, multiply the sum by the num ber of terms, and divide by 2.

EXAMPLES:

1. The extremes being 3 and 19, and the number of terms 9; required the sum of the terms?

2. It is required to find the number of all the strokes a common clock strikes in one whole revolution of the index, or in 12 hours?

Ans. 78.

Ans. 5 oz of 15, of 17, and of 18 caracts fine, and 25 oz of 22 caracts fine*.

Ex. 2. A vintner has wine at 48, at 5s, at 58 6d, and at 6s a gallon; and he would make a mixture of 18 gallons, so that it might be afforded at 58 4d per gallon; how much of each sort must he take?

Ans. 3 gal. at 4s, 3 at 5s, 6 at 58 6d, and 6 at 6s.

A great number of questions might be here given relating to the specific gravities of metals, &c. but one of the most curious may here suffice.

Hiero, king of Syracuse, gave orders for a crown to be made entirely of pure gold; but suspecting the workman had debased it by mixing it with silver or copper, he recommended the discovery of the fraud to the famous Archimedes, and desired to know the exact quantity of alloy in the crown.

Archimedes, in order to detect the imposition, procured two other masses, the one of pure gold, the other of silver or copper, and each of the same weight with the former; and by putting each separately into a vessel full of water, the quantity of water expelled by them determined their specific gravities; from which, and their given weights, the exact quantities of gold and alloy in the crown may be determined.

Suppose the weight of each crown to be 101b, and that the water expelled by the copper or silver was 921b, by the gold 52lb, and by the compound crown 641b; what will be the quantities of gold and alloy in the crown?

The rates of the simples are 92 and 52, and of the compound 64; therefore

9228 of gold

And the sum of these is 12+28 40, which should have been

but 10; therefore by the Rule,

=

40: 10: 12: 3lb of copper
40 10 28 71b of gold

the answer.

RULE

EXAMPLES.

1. Given the least term 3, the common difference 2, of an arithmetical series of 9 terms; to find the greatest term, and the sum of the series.

2. If the greatest term be 70, the common difference 3, and the number of terms 21, what is the least term, and the sum of the series?

Ans. The least term is 10, and the sum is 840.

3. A debt can be discharged in a year, by paying 1 shilling the first week, 3 shillings the second, and so on, always 2 shillings more every week; what is the debt, and what will the last payment be?

Ans. The last payment will be 51 38, and the debt is 1357 48.

PROBLEM IV.

To find an Arithmetical Mean Proportional between Two Given

Terms.

ADD the two given extremes or terms together, and take half their sum for the arithmetical mean required.

EXAMPLE.

To find an arithmetical mean between the two numbers 4 and 14.

Here
14
4

2) 18

Ans.

9 the mean required.

PROBLEM

PROBLEM V.

To find two Arithmetical Means between Two Given Extremes.

SUBTRACT the less extreme from the greater, and divide the difference by 3, so will the quotient be the common difference; which being continually added to the less extreme, or taken from the greater, gives the means.

EXAMPLE.

To find two arithmetical means between 2 and 8.

To find any Number of Arithmetical Means between Two Given

Terms or Extremes.

SUBTRACT the less extreme from the greater, and divide the difference by 1 more than the number of means required to be found, which will give the common difference; then this being added continually to the least term, or subtracted from the greatest, will give the mean terms required.

EXAMPLE.

To find five arithmetical means between 2 and 14.

Here 14
2

6) 12 Then by adding this com. dif. continually,
the means are found 4, 6, 8, 10, 12.

com. dif. 2

See more of Arithmetical progression in the Algebra.

GEOMETRICAL

Then, 50: 120 :: 60: 144, the Answer.

per question.

2. What number is that, which being multiplied by 7, and the product divided by 6, the quotient may be 21? Ans. 18.

3. What number is that, which being increased by }, }, and of itself, the sum shall be 75? Ans. 36.

4. A general, after sending out a foraging and of his men, had yet remaining 1000; what number had he in command? Ans. 6000.

5. A gentleman distributed 52 pence among a number of poor people, consisting of men, women, and children; to each man he gave 6d, to each woman 4d, and to each child 2d moreover there were twice as many women as men, and thrice as many children as women. How many were there of each? Ans. 2 men, 4 women, and 12 children?

6. One being asked his age, said, if of the years I have lived, be multiplied by 7, and of them be added to the product, the sum will be 219. What was his age?

Ans. 45 years.

DOUBLE