EVOLUTION OF SURDS. 92. RULE. Extract the required root of the coefficient, and then multiply the fractional exponent of the surd by the fractional exponent of the root. EXAMPLE. Extract the square root of 9 ab. Here the square root of 9 is 3, and the fractional exponent of the surd is, which we are to multiply by the exponent of the root, which gives ; hence the quantity sought is 3(ab). 1. Extract the square root of 9/3. 2. Extract the square root of 36/2. 3. Extract the cube root of 8.2/5. 4. Extract the cube root of 27/7. Ans. 3/3 Ans. 6/2 Ans. 2 Ans. 3×78. 5. Extract the fourth root of 643/4. Ans. 2× (256) 11⁄2 EQUATIONS CONTAINING SURDS, ETC. 93. In equations containing surds, before the solution can be effected, the surds must be cleared away; to effect this, transpose all the terms which do not contain surds to one side of the equation, and the surds to the other, then raise both sides to a power denoted by the index of the surd, and if there was only one term containing a surd, the surd will be cleared away, if there be more than one, the operation must be repeated. If an equation appear under the form xa√xb, or x2axc, it can be solved as an adfected quadratic, by solving first for the power in the second term, and then for the quantity itself. EXAMPLE. Given √x+9=√x+1. Squaring both we have x+9=x+2√x+1. 5 EXAMPLE 2. Given x3+x2=72, to find the value of x. Here the equation comes under the form x2+x=c, since the exponent of the first term is double its power in the second; hence we must solve for x. The operation will be as follows: x3+x2=72. x2-8 or-9, by transposition. and x64 or 81, by squaring. hence x=4 or 33/3, by extracting the cube root. EXERCISES. 1. Given 3x+4=5, to find x. 2. Given √4+5x=2+√3x; to find æ. 3. Given /2x+10+4=8; to find x. 3 4. Given +5=7; to find æ. Ans, x=7. Ans. x=12. Ans. x=27. Ans. x6. 5. Given √4x+17+6√x+2=8√x+3; to find x. 6. Given √x+√x-7= Ans. x 16. 11. Given √√√ax, to find x. Ans. x= 12. Given √x+a+√a—x=b, to find x. 13. b a Ans. ! (4a—b2) &. x= 2 Given√1+x√x2+12=1+x, to find x. Ans.x=2. 1. 18. Given x+2x=24, to find x. 19. Given x2-—2x2=x, to find x. Ans. x16, or 36. Ans. x 4, or 1. 20. Given x+x=6, to find x. Ans. x=32, or -243. 21. Given 3x-2x-133, to find x. Ans. x=49, or 361. 22. Given (x+12)*+(x+12)=6, to find x. Ans. x=4, or 69. ARITHMETICAL PROGRESSION. 94. Definition. If a series of quantities increase or decrease, by the constant addition or subtraction of the same quantity, then the quantities are said to be in arithmetical progression; and the quantity thus added or subtracted is called the common difference. Thus, 2, 5, 8, 11, &c., is an increasing arithmetical progression, where the common difference is 3; and 19, 17, 15, 13, &c., is a decreasing arithmetical progression, in which the common difference is 2. In general, if a represent the first term, l the last term, d the common difference, n the number of terms, and s the sum of all the terms, the progression may be thus expressed : Where the first represents an increasing, and the second a decreasing series. In each of the series it may be observed, that the coefficient of d is always one less than the number of that term in the series; hence the nth or last term is equal to a+n-ld, that is, la+n-ld. 95. To find s; observe that the series may be written by beginning with the last term and subtracting d; thus 7, l—d, l—2d, l—3d, l—4d, l—n—ld, where it is obvious, that nida; writing the series then in both forms, and then adding; thus, s=a+a+d+a+2d+a+3d...+a+n—ld, where the number of terms on the second side is evidently n "; :. 28=(a+1)n, and s=(a+1). In which substitut ing instead of its value a+ (n-1)d, we obtain s= From the equations found above, namely, l=a+n—1d, and s={2a+n—1d}}, by substitution and reduction the following theorems may be deduced, from which it will appear, that any three of the five quantities, a, d, l, n, s, being given, the remaining two may be found. 19th, 8= 12-a2 2+1+a). 20th, s={21—n—1d}". 2d The above 20 theorems are sufficient for the solution of any question that can be proposed in arithmetical progression; the pupil should deduce the theorems from the two given equations, it being one of the best exercises in literal analysis that can be given. EXERCISES. 1. The first term of an arithmetical series is 5, the common difference 4, and the number of terms 12; find the last term, and the sum of the series. Apply Theorem 5th to find l, and Theorem 18th to find s. Ans. 149, and s=324. 2. Given the first term 3, the last term 51, and the common difference 2, to find the number of terms and the sum of the series. Substitute in Theorem 14th to find n, and in Theorem 19th to find s. Ans. n 25, and s= :675. 3. Given the sum of an arithmetical series 12,100, the first term 1, and the common difference 2. Find the last term, and the number of terms. Substitute in Theorem 8th to find l, and in Theorem 15th to find n. Ans. 7 219, and n=110. 4. A person was forty years in business, and spent the first year L.80, and increased his expenditure annually by L.4. What did he spend the last year, and how much during the whole forty? Ans. He spent the last year L.236, and during the whole forty years L.6320. 5. The first term of a decreasing arithmetical series is 9, the common difference, and the number of terms 21; find the sum of the series. Ans. 119. 6. A man is to receive L.300 at twelve several payments, each payment to exceed the former by L.4. He is willing to bestow the first payment on any one that can tell him what it is. What must the arithmetician have for his pains? Ans. L.3. GEOMETRICAL PROGRESSION. 96. Definition. If a series of terms be such that each can be produced from the immediately preceding one, by multiplying by the same number, the series is called a geometrical progression; and the series is an increasing or decreasing one, according as the multiplier is greater or less than unity. Thus, 1, 2, 4, 8, 16, &c., is an increasing geometrical series, where the common multiplier is 2; and 243, 81, 27, 9, 3, &c., is a decreasing geometrical series, in which the common multiplier is The common multiplier is called the ratio, and is commonly represented by r; and if a be put for the first term, the general representation of a geometrical series will be the following: a, ar, ar2, ar3, ar1, &c.; and if n be put for the number of terms, and s for the sum of the series, we will have 8=a+ar+ar2+ar3 +art+...arn−1. Multiply both sides of (1) by r, and it becomes srar+ar2+ar+ar1+...arn-1+arn. (1.) (2.) |