Then take the sum of The assumed power multiplied by its index added to 1; And the sum of The assumed power multiplied by the index lessened by 1; And the given number multiplied by the index added to 1 Then say, by the Rule of Proportion, As the first of those sums, Is to the second, So is the assumed root, To the required root, nearly. And if this root be taken for the assumed root, and the operation repeated, a nearer approximation will be obtained; and so on. Examples of the Cube Root. 1. Required the 3d. or cube root of 184? Assume 6 for the root, whose cube is 216, the assumed power. Therefore, As the sum of 216 × 4 and 184 × 2, Is to the sum of 216 × 2 and 184 x 4; So is the assumed root 6, To the root, nearly. Or dividing the two first terms of the proportion by 2 we have (96.) As the sum of216 x 2 and 184, Is to the sum of 216 and 184 X 2; So is 6, To the root, nearly. In words, As twice the assumed cube added to the given number, To the required root, nearly. 77 Now taking 5-7 for the assumed root, its cube is 185 193 the assumed As 554-386 553.193 :: 57: 5.687734 root, which is true in the last decimal. Now take 068921 the cube of 41 for the second assumed cube, As 207842 208921: 41: 4121285 root, true to the last figure. For 4121285306999998 (retaining 8 places of decimals only) which is less than ⚫00000002 short of the truth. 119. To extract the cube root of a Vulgar Fraction. Reduce it to its lowest terms: then the roots of the numerator and denominator will form the fractional root required, Thus the cube root of is 18. And the cube root of 24 is 3; for 24 = whose root is 3. But when the terms of the fraction are not perfect cubes, let them be multiplied by the square of the denominator, then extract the root of the new fraction for the root required. Thus, suppose the cube root of is required. Or the fraction may be reduced to a decimal. And mixt numbers are prepared as in extracting the square root. 120. The Biquadratic or 4th root is obtained by extracting the square root, and then extracting the square root of that root. Thus the 4th root of 6561 is 9. For the square root of 6561 is 81 whose square root is 9. Let the 5th root of 27 be required, Assume 2 for the root; then its 5th power is 32. And the index 5 added to 1, and lessened by 1 give 6 and 4. Then 32 X 6 = 192 27 X 4 = 108 Sum 500 32 X 4 = 128 As 300: 290 :: 2 : 193, root nearly, the first approximation. Now assume 1.93 for the root; then its 5th. power, or the assumed power is 26-778 (retaining 3 places of decimals only). 26778 x 6 = 160-668 27 X 4 = 108 Sum 268-668 26.778 x 4 = 107.112 As 268-663: 269.112:: 1.93: 1-933181 the root true to the last figure. OF ARITHMETICAL PROPORTION AND PROGRESSION. 121. WHEN four numbers have a common difference they are said to be in continued arithmetical proportion. But if the difference of the first and second is equal to the difference of the third and fourth, but not to that between the second and third, it is called discontinued proportion. 2, 4, 6, 8, continued proportion. 2, 4, 7, 9, discontinued proportion. 122. A series or rank of the first kind form a progression: 1, 2, 3, 4, 5, 6, &c. 0, 1⁄2, 1, 1, 2, 24, &c. ascending series or progressions. 32, 29, 26, 23, 20, 17, &c. } descending progressions. 123. The first and last numbers or terms are called the extremes; and the others between them the means. Thus 1 and 6 are the extremes; and, 2, 3, 4, 5, the means of the rank 1, 2, 3, 4, 5, 6. 124. It is evident from the nature of the progressions, that the double of any term is equal to the sum of the two adjacent terms, or to the sum of any two terms equidistant from it. Thus in the rank 1, 2, 3, 4, 5, 6, &c. twice 43+ 5 = 2 + 6. 125. Hence if three numbers are in arithmetical proportion, twice the mean is equal to the sum of the two extremes. Thus, if the three numbers are 104, 10, 94, Then 10 x 210194. 126. And when 4 numbers are in arithmetical proportion, the sum of the two means is equal to that of the extremes. Thus if 32, 29, 20, 17, are the 4 numbers, Then 29 420 = 32 + 17. 127. Since the terms of an arithmetical progression are found by continually adding or subtracting the common difference; if the difference, twice the difference, three times the difference, &c. be added to the first term, the several sums will give an ascending series; or subtracted, a descending one. Thus the terms of the progression 3, 5, 7, 9, 11, &c. having the common difference 2, will be 3, 32, 3+ 4, 3 + 6, 3 +8, &c. And the terms of the series 6, 51, 5, 41, 4, &c. where the common difference is 1, is 6, 61, 6 — 1, 6 — 1 1⁄2, 6-2, &c. 128. Consequently when the first and last terms are given, if their difference be divided by the number of terms lessened by 1, the quotient will be the common difference of the terms. For example, let the first term be 2, the last 20, and the number of terms 7: Then 20 2 18 the difference, which divided by 6 (or 7—1) gives 3 for the common difference of the terms. And the progression will be 2, 5, 8, 11, 14, 17, 20. 129. In this manner we can find any proposed number of arithmetical means between two given numbers; or interpose any number of terms between two given extremes. For example, let 9 arithmetical means be found between 1 and 2. Now the whole number of terms being 11, that nuraber lessened by 1 is 10: And 2 -11 the difference of the extremes,, which divided by 10 gives the common difference of the terms, |