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6. What is the 9th root of 1.577635?
7. What is the 12th root of 16.3939? 8. What is the 18th root of 104.9617? Ans. 1.2950+.
444. When the index of the required root is prime, or contains any other factor than 2 or 3.
To extract any root of a number is to separate the number into as many equal factors as there are units in the index of the required root; and it will be found that if by any means we can separate a number into factors nearly equal to each other, the average of these factors, or their sum divided by the number of factors, will be nearly equal to the root indicated by the number of factors.
1. What is the 7th root of 308?
SOLUTION. We first
2.59 +2.04 = 4.63.
4.63 ÷ 2 = 2.31, assumed root.
308151.93 = 2.0272+.
2.31 × 6+2.0272 = 15.8872.
15.8872 ÷ 7 = 2.2696, 1st approx.
find the 6th root, and
also the 8th root of 308; and since the 7th root must be less than the former and greater than the latter, we take the average of the two, or one half of their sums, 2.31, and call it the assumed root. We next raise the assumed root, 2.31, to the 6th power, and divide the given number, 308, by the result, and obtain 2.0272+ for a quotient; we thus separate 308 into 7 factors, 6 of which are equal to 2.31, and the other is 2.0272. As these 7 factors are nearly equal to each other, the average of them all must be a near approximation to the 7th root. Multiplying the 2.31 by 6, adding the 2.0272 to the product, and dividing this result by 7, we find the average to be 2.2696, which is the first approximation to the required root. We next divide 308 by the 6th power of 2.2696, and PRAC. AR.- -24
2.2696 × 6 +2.253452 = 15.871052. 15.871052+7=2.267293, 2d approx.
obtain 2.253452+ for a quotient; and we thus separate the given number into 7 factors, 6 of which are each equal to 2.2696, and the other is 2.253452. Finding the average of these factors, as in the former steps, we have 2.267293, which is the 7th root of the given number, correct to 5 decimal places.
RULE.-I. Find by trial some number nearly equal to the required root, and call this the assumed root.
II. Divide the given number by that power of the assumed root denoted by the index of the required root less 1: to this quotient add as many times the assumed root as there are units in the index of the required root less 1, and divide the amount by the index of the required root. The result will be the first approximate root required.
III. Take the last approximation for the assumed root, with which proceed as with the former, and thus continue till the required root is obtained to a sufficient degree of
If the index of the required root contains the factors 2 or 3, we may first extract the square or cube root as many times, successively, as these factors are found in the index, after which we must extract that root of the result which is denoted by the remaining factor of the index. Thus, if the 15th root were required, we should first find the cube root, then the 5th root of this result.
445. An Arithmetical Progression, or Series, is a series of numbers increasing or decreasing by a constant common difference.
Thus, 3, 5, 7, 9, 11, etc., is an arithmetical progression with an ascending series, and 13, 10, 7, 4, etc., is an arithmetical progression with a descending series.
446. The Terms of a series are the numbers of which it is composed.
447. The Extremes are the first and last terms. 448. The Means are the intermediate terms.
449. The Common Difference is the difference between any two adjacent terms.
450. There are five parts in an arithmetical series, any three of which being given, the other two may be found. They are as follows: the first term, last term, common difference, number of terms, and sum of all the
451. To find one of the extremes when the other, the common difference, and number of terms are given.
1. Let 2 be the first term of an ascending series of 8 terms, and 3 the common difference. Find the 8th term. The series will be written 2, 5, 8, 11, 14, etc., or analyzed, thus, 2, 2 +3, 2+3+3,2+3+3+3, 2+3+3 +3 +3.
3 x7 = 21.
SOLUTION. Since the series will be 2, 5, 8, 11, 14, or analyzed 2, 2 + 3, 2+3+3, 2 + 3 + 3 + 3, 2+3 +3 +3 + 3, etc., we see that in an ascending series, we obtain the second term by adding the common difference once to the first term; the third term, by adding the common difference twice to the first term; the fourth term, by adding the common difference three times to the first term, etc.; and, in general, we obtain any term by adding the common difference as many times to the first term as there are terms less one. Hence, the eighth term will be equal to the first term, 2, plus 7 × 3 = 23, Ans.
1st term, 2+21=23, 8th term.
RULE. Multiply the common difference by the number of terms less 1, and add the product to the first term, if the series is ascending, and subtract it if the series is descending.
2. The first term of an ascending series is 4, mon difference 3, and the number of terms 19. the last term?
3. What is the 13th term of a descending series whose first term is 75, and common difference 5?
4. A boy bought 18 hens, paying 2 cents for the first, 5 cents for the second, and 8 cents for the third, in arithmetical progression. What did he pay for the last hen?
5. What is the 40th term of the series, 4, 1, 14, etc.? Ans. 101. 6. A man travels 9 days; the first day he goes 20 miles, the second 25 miles, increasing 5 miles each day. How far does he travel the last day of his journey?
Ans. 60 miles. 7. What is the amount of $100, at 7%, for 45 years? $100 + $7 x 45 $415, Ans.
8. The first term of an arithmetical progression is 5, the common difference 4, and the number of terms 8. What is the last term?
452. To find the common difference when the extremes and number of terms are given.
1. The first term is 2, the last term 23, and the number of terms 8. What is the common difference?
SOLUTION. Referring to the series, 2, 5,
8, 11, 14, analyzed in 451, we readily see that, term, we have left the
by subtracting the first term from any common difference taken as many times as there are terms less 1. Thus, by taking away 2 in the fifth term, 2 + 3 + 3 + 3 + 3, we have left 12, which is 4 times 3. Hence, we divide the difference between 2 and 23, 21, by 8-1=7, and find the common difference 3.
Divide the difference of the extremes by the number of terms less 1.
2. The first term is 2, the last term is 17, and the number of terms is 6. What is the common difference? 3. A man has seven children whose ages are in arith metical progression; the youngest is 2 years old, and the eldest 14. What is the common difference of their ages? Ans. 2 years.
4. The extremes of an arithmetical series are 1 and 501, and the number of terms is 34. What is the common difference?
5. An invalid commenced to walk for exercise, increasing the distance daily by a common difference; the first day he walked 3 miles, and the 14th day 9 miles. How many miles did he walk each day?
When we have found the common difference, we may add it once, twice, etc., to the first term, and we have the series, and consequently the means.
Ans. 3, 31, 4, 41, 5, 51, etc.
6. The first term of an arithmetical progression is 5, the last term 54, and the number of terms 8. What is the common difference?