CHAPTER XXII. LITERAL EXPONENTS. Letters are sometimes placed as exponents instead of numbers, but as they represent numbers, they are subject to the same rules. Thus a" × a”÷a”+”, am÷a”÷a”-”, (a)"a", and "a"a". EXERCISE A. Express the following products, &c., in their simplest forms :-a2m × a3m, bên ÷ b3n, c3m × c3n, xTM-′′ × x”, y"+"÷y", zTM+" × 2"-", 2"+"÷÷2m-", (dm)3, (62m)3n, √(am) and (m). 1. Cube a"-b". 2m zm Ans. am-3amb” +3amb2 — b3n 2. Multiply 7y3 +524 by 7y3r — 5249. Ans. 49yor-25289. 3m C. Multiply am+am+a+1 by am-a3m. 3. Divide 6c3-13c2+6c" by 3c-2. Ans. 2c2-3c. D. Divide 22 - x2my" — 2xy31-8y3 by x-2y". 3m 4. Divide a"-16a" by a2 + 2a”. E. Divide 252+50≈2-100≈1+200 by 522-1022+10. Ans. an-2a2+4a" -8. c2m+2cmm + 22m and 2m 22m to their lowest terms. x + 3x2 + 4x” + 2' 7. Find the square root of 4 - 2x2m2 + y1n + 2x22 -2y2n22P+zAP. H. Determine (an-4a5-2a+12a3+9a). 8. Given 2+2ax"= b2, to find x. Ans. x=√(a2 + b2)—a}. K. Given 24-322m+ 4 = 0, to find z. CHAPTER XXIII. (21.)* ARITHMETICAL PROGRESSIONS. DEFINITIONS. If a series of quantities increase or decrease by equal differences, it is called an Arithmetical Progression,as 7, 9, 11, 13, 15, 17; or 16x, 14x, 12x, 10x, 8x. The difference between any two adjacent terms is called the Common Difference. If a represent the first term, d the common difference, and n the number of terms, any arithmetical progression will be represented thus: EXERCISE. Fill up the blanks in the following arith In any Arithmetical Progression, the Sum of the two extreme Terms is equal to the Sum of any two Terms equidistant from the Extremes, or to Double the middle Term when the Number of Terms is odd. DEM. If a be put for the first term and for the last, the common difference being d, the progression will be expressed thus:— a, ad, a±2d,...l 2d, ld, l. The same progression reversed, is l, ld, l = 2d,............a ± 2d, a ±d, a. • The number enclosed in brackets indicates the Chapter in the third edition. Adding together these two forms of the progression in the order of the terms, we have the sum of every pair =a+l; and, if the number of terms is odd, the middle term may be expressed in the one line by a md, and in the other by md, the sum of the two being also a+l. NOTE. Any part of an arithmetical progression is, of course, an arithmetical progression. EXERCISE A. Fill up the blanks in the progression, 7, 19. 1. What is the middle term of an arithmetical progression of eleven terms, having its first term 156a, and its last term 246a+ 10d? Ans. 201a+5d. B. The second term of an arithmetical progression is x+10y; and the sixth term, x+38y. What is the fourth term? PROBLEM I. Having given the common Difference of an Arithmetical Progression and the first Term, to find any other Term (say the nth). RULE. Multiply the common difference by 1 less than the number of the term required (i. e. by n-1), and add the product to the first term, or subtract it from it, according as the progression increases or decreases. DEM. The Rule is derived immediately from the general form of a progression, employed in illustrating the Definition, viz. a, a±d, a± 2d, a ± 3d, ............a ± (n − 1)d, in which we observe that the 2d term is a ±1d; the 3d, a2d; the 4th, a± 3d; and consequently, in general, the nth, a ±(n-1)d. EXAMPLE. What is the twentieth term of the progression 3a, 5a, 7a, &c.? 3a+2a × 19=41a. Ans. EXERCISE 1. What is the fiftieth term of the arithmetical progression, 1, 3, 5, &c.? Ans. 99. 2. What is the tenth term of the progression 30c, 27c, 24c, &c.? Ans. 3c. A. What is the eighth term of the progression 17, 32, 47, &c.? B. What is the eleventh term of the progression 25a, 20a, 15a, &c.? C. What is the twelfth term of the series, a, a+5, a+10, &c. ? D. Compute the sixteenth term of the progression x+4y, x+2y, x, x-2y, &c. PROBLEM II. Having given any two Terms of an Arithmetical Progression, and the Number of intermediate Terms, to find the Common Difference. RULE. Divide the difference of the two given terms by 1 more than the number of intermediate terms: the quotient will be the common difference. DEM. If the first of the two given terms be called a; the last, ; the com. dif., d; and the whole number of terms from a to 7 inclusive, n, the number of intermediate terms being n-2; then, by Pr. I., la±(n-1)d. ...(n-1)d=1-a, and d(-a)÷(n-1), (n-1) being 1 more than the number of intermediate terms. EXERCISE 1. The first term of an arithmetical progression is 1; and the tenth, 46. What is the common difference? Ans. 5. A. The fifth term is 63, and the eighth 96. What is the common difference? B. The second term is x+10y; and the sixth x+38y. Determine the common difference. C. Complete the progression, 85, —‚—‚—‚ by inserting the blank terms. PROBLEM III. To find the Sum of an Arithmetical Progression. 175, RULE. Multiply half the sum of the first term and the last by the number of terms. DEM. Expressing the progression, and reversing it as in the Theorem above, and putting also s for the sum, and n for the number of terms, we have First, s a +(a±d) + (a ±2d)....+ (l + 2d) + (l + d) + ↳ and, sl+(d) + (l = 2d)....+(a±2d)+(a±d)+a. = Adding the two equations together, and observing that, when the upper sign of the two is used in the first equation, the upper is also used in the lower, and vice versa, we have 2s=(a+1)+(a+1)+(a+1), to n terms, =(a+1) × n. (a+1) x n a+1 2 or x n. 2 This will perhaps be more easily understood, if we take a particular case, as the progression 7, 10, 13, &c., to 14 Then terms. s 7+10+13....... +40 +43 + 46. and s46+43 + 40.........+ 13 + 10 + 7. ..2s=53 +53 +53.........+53 +53 +53; or 2s 53 x 14. 3х 2. Find the sum of 21 terms of the series, x, 2.x, 5x 2' 2' Ans. 126x. &c. C. Sum the series 5a, 5a + 3b, 5a + 6b, &c., to seven terms. 3. A young man, engaging as a farm servant, asked ten pounds a year as his wages. His employer refused to give so much at first, but offered him 5 guineas for the first year, to be raised 15 shillings every year that he remained in his place. He remained 40 years on these terms. What were his last year's wages, and what did he earn during the whole time? Ans. £34, 10s.; and £795. D. How far would a person walk before he could pick up 100 pebbles, placed in a straight line, a yard apart, and bring them one by one to the place where he stood, which was in the same straight line, and one yard from the first pebble in the row? |