4. Multiply this root figure by the part of the root previously obtained, annex one cipher and multiply this result by three; add the last product and the square of the last root figure to the trial divisor, and the SUM will be the TRUE DIVISOR.

5. Multiply the true divisor by the last root figure, subtract the product from the dividend, and to the remainder annex the next period for a new dividend.

6. Find a new trial divisor, and proceed as before, until all the periods have been employed.

NOTE 1. The notes in Art. 357, with slight modifications, are equally applicable here.

NOTE 2. If the root consists of three figures it is plain that the cube, as completed in Fig. 4, must be enlarged just as Fig. 1 has already been enlarged. Hence, the new trial divisor will consist of 3 faces of Fig. 4; but the true divisor already found is the sum of the significant figures in these 3 faces, except one face each of rr, xx, and zz, and two faces of the little corner cube, nnn; moreover, the number directly above the true divisor (in the operation) represents one face of nnn, and the number above that represents the sum of one face each of the 3 long corner blocks, rr, xx, and zz; hence, to find the next trial divisor, we have only to add the true divisor already found to TWICE the number above it, and ONCE the number above THAT, and to the sum annex two ciphers. When there are many root figures this process is shorter than to square so much of the root as has been found, annex two ciphers, and multiply by 3, as directed in the 3d paragraph of the rule. Ex. 2. What is the cube root of 21024576?

The 1st trial divisor is contained 10 times in the dividend, yet the root figure is only 7. The true root figure can never exceed 9, and must in all cases be found by trial.

Squaring 20 gives the same result as squaring 2 and annexing 00, as directed in the rule, 3d paragraph.

In this example, the 1st trial divisor, 4800, is larger than the 1st dividend, 3917; .. we annex 0 to the root, 00 to the 1st trial divisor for the 2d trial divisor, and bring down the next period to complete a new dividend. The rule, followed literally, will give the same result.

NOTE 3. Prepare fractions and mixed numbers as directed in square root (Art. 359).

What is the value of the following expressions :

3

4. 2803221?

5. 3/3176523 ? 6.382657176? 7.8024024008 ?

8. 3/387420489 ?

3

Ans. 141. 11. /36.926037? Ans. 3.33.

10. 3/5? Ans. 1.709. 17. 3/435?

APPLICATION OF THE CUBE ROOT.

366. Bodies which are of precisely the same form are similar to each other, and the solid contents of similar bodies are to each other as the cubes of their corresponding lines, and conversely, the corresponding lines are to each other as the cube roots of the contents.

Ex. 1. If an iron ball 5 inches in diameter weighs 16 pounds, what is the weight of a ball 30 inches in diameter?

53 : 303 :: 16 : Ans., or 13 : 63 :: 16 : Ans.; i. e. 1: 216 :: 16lb.: 3456 lb., Ans.

365. What is Note 1? Note 2? Explain Ex. 2. Ex. 3. What is Note 3? 366. What are similar bodies? The ratio of the contents of similar bodies?

2. If a ball 6 inches in diameter weighs 27 pounds, what is the diameter of a ball that weighs 64 pounds?

3/27:3/64 :: 6in.: Ans.; i. e. 3: 4 :: 6in. : 8in., Ans. 3. How many bullets of an inch in diameter will be required to make a ball 1 inch in diameter?

4. If a globe of gold 1 inch in diameter is worth $100, what is the diameter of a globe worth $6400?

5. Suppose the diameter of the earth is 7912 miles, and that it takes 1404928 bodies like the earth to make one as large as the sun, what is the diameter of the sun?

6. A bin is 8 feet long, 4 feet wide, and 2 feet deep; what is the edge of a cubical box that will hold the same quantity of grain?

7. If a stack of hay 24 feet high weighs 27 tons, what is the hight of a similar stack which weighs 8 tons? Ans. 16ft.

8. If a bell 4 inches high, 3 inches in diameter, and of an inch thick weighs 1lb., what are the dimensions of a similar bell that weighs 27 lb. ?

9. If a loaf of sugar 10 inches high weighs 8 lb., what is the hight of a similar loaf weighing 1lb.?

ARITHMETICAL PROGRESSION.

367. Any series of numbers increasing or decreasing by a common difference is in ARITHMETICAL PROGRESSION;

thus, 2, 5, 8, 11, 14, 17, etc. is an ascending series, and 35, 30, 25, 20, 15, 10, etc. is a descending series. The several numbers forming a series are called TERMS; the first and last terms, EXTREMES; the others, MEANS. The difference between two successive terms is the COMMON DIFFERENCE.

367. When is a series of numbers in Arithmetical Frogression? How many kinds of series? What? What are the Terms of a series?

In an arithmetical series 5 particulars claim special attention, viz. the first term, last term, common difference, number of terms, and sum of all the terms; and these are so related to each other that if any three of them are given the other two can be found. 368. In an ascending series, let 6 be the first term and 5 the common difference;

Then

6: = 1st term. 6+5=11 = 2d term.

6+5+5 6+2 × 516=3d term. 6+5+5+5=6+3x5=214th term.

Thus we see that, in an ascending series, the second term is found by adding the common difference once to the first term; the third term, by adding the common difference twice to the first term, etc.

A similar explanation may be given when the series is descending. Hence,

369. PROBLEM 1. To find the last term, the first term, common difference, and number of terms being given :

RULE. Multiply the common difference by the number of terms less 1; add the product to the first term if the series is ascending, or subtract the product from the first term if the series is descending, and the sum or difference will be the term sought.

Ex. 1. If the first term of an ascending series is 5, the common difference 4, and the number of terms 7, what is the last term? 5 + 6 × 4 = 29, Ans.

2. The first term of a descending series is 47 and the common difference 8; what is the 6th term?

47-5 X 87, Ans. 3. What is the amount of $100, at 6 per cent., simple interest, for 25 years?

367. What are the Extremes of a series? Means? Common Difference? How many particulars claim special attention? What are they? How many of them must be given? 368. How is an ascending series formed?

ing series? 369. Object of Problem 1? Rule?

How a descend

370. PROBLEM 2. To find the common difference, the extremes and number of terms being given.

By inspecting the formation of the series in Art. 368, it will be seen that the difference between the extremes is equal to the common difference multiplied by 1 less than the number of terms; e. g. the difference between the 1st and 4th terms (21 6=15), is the sum of 3 equal additions; .. this difference, divided by 3 (15 ÷ 3 = 5), gives one of these additions, i. e. the common difference. Hence,

RULE. Divide the difference of the extremes by the number of terms less one, and the quotient will be the common difference.

Ex. 1. The extremes of an arithmetical series are 3 and 38, and the number of terms is 8; what is the common difference? 38 3 35

= 5, Ans.

2. A man has 6 sons whose ages form an arithmetical series; the youngest is 2 years old and the oldest 22; what is the difference of their ages ? Ans. 4 yr.

3. The amount of $100 at simple interest for 10 years is $160; what is the rate per cent.?

371. PROBLEM 3. To find the number of terms, the extremes and common difference being given.

By Art. 368 it is evident that the difference of the extremes is the common difference multiplied by one less than the number of terms. Hence, conversely,

RULE. Divide the difference of the extremes by the common difference, and the quotient, increased by 1, is the number of

terms.

Ex. 1. The extremes of an arithmetical series are 3 and 31 and the common difference is 4; what is the number of terms?

2. The common difference in the ages of the children in a family is 2 years; the youngest is 1 year old and the oldest 19; how many children in the family?