3 18. Given x+2 x=24, to find X. Ans. x=16, or 36. 19. Given x2–2x2 = x, to find x. x Ans. x=4, or 1. 20. Given x3 4x3 =6, to find x. +x3 =6, to find X. Ans. x=32, or —243. 21. Given 3x-22=133, to find X. Ans. x=49, or 36 1 22. Given (x+12)+(x+12)+=6, to find x. Ans. x=4, or 69. 9 Ist term, zd, 4th, or a, 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, 7 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 :2d, last. a, atd, a+2d, at 3d, le ad, a-2d, a-3d, 1. 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 atid, that is, l=a+n–14. 95. To find s; observe that the series may be written by beginning with the last term and subtracting d; thus l, -d, 1-20, 1-3d, 1-4d, 1-n-ld, where it is obvious, that lon-d=a; writing the series then in both forms, and then adding; thus, 8=a+a+d+a+20 +a+3d... ta+n-1d, 2s=(a+7)+(a+1)+(a +1)+(a+l)... +(a+), where the number of terms on the second side is evidently ? ; :: 25=(a + 1)n, and s=(a+1). In which substitut n S n 3d, a= n 2s a. n 2 ing instead of l its value a+(n-1)d, we obtain s= (2a+n—12) 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 three of the five quantities, a, d, l, n, s, being given, the remain. ing two may be found. THEOREMS. 1st, a=l-n-ld. 2d, a= -d. 28 -1. 4th, a=iv (21+d)2—8ds+id. 5th, l=a+n-ld. 6th, l= n-1 7th, l=+ d. 8th, l=iv (20—d)? +8d8-4d. Ha 25—2an 9th, d= 10th, d= n(n-1) la 12th, dr Х 28-(a+1) 2s 13th, n= 14th, n= {V(2a—d)? +8ds+d—2a}. 2d 16th, n= 21+- (21+d)-8d8 2d atl 17th, s= 18th, s={2a +n—10}.. 12_a2 19th, s= +}(l+a). 20th, 20th, s={2——ld. 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. In-8 Ilth, d= (1+a)(La) n n +1. 15th, n= n. 2 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. I=49, 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. l=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 Ans. L.3. for his pains ? 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 Jeometrical 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, ara, ary, art, &c.; and if n be put for the number of terms, and s for the sum of the series, we will have szatartara tar3 + ant t...am-1. (1.) Multiply both sides of (1) by r, and it becomes sr=ar+ar+ar+ari +...am-" +ar", (2.) Subtract (1) from (2), and there remains sr-sara. The other terms destroy one another. Hence (r-l)s=a(ru_1). a(1) by dividing by pl. (3.) This is the formula for s, when r is greater than unity; but if r is less than unity, the first term of the series will be the greatest, and the proper expression for s is obtained by subtracting (2) from (1), which gives S-sr=ą-ar". Hence (1—r)s=a(1—2ore), a(1- (4.) (1---) If now we represent the last term byl, it is (1) evident that l-am-1. From these two equations, namely, l=ap-), and s= a(1) a(14"), the following theorems may be derrived. (n-1)s 1st, as 2d, a= > 1 3d, a=lr(r-118 4th, r= gong-1 llth, l= X 8. pl The above theorems are given as exercises in literal analysis, and should all be deduced from the 6th and 9th, which were previously proved. When ~ is less than 1, the term yn may be rendered less than any quantity that can be named, however small, by taking n sufficiently great; so that (4) in the case of n=infinity, will become 1 which is the expression for the sum of a decreasing geometrical series continued to infinity, a 8= EXERCISES. 1. Given the first term of a geometrical series 1, the common ratio 2, and the number of terms 10, to find the last term and the sum of the series. Substitute in Theorems 9th and 6th, and we have Ans, l=512, s=1023. 2. The sum of a geometrical progression, whose first term is l, and last term 128, is 255. What is the common ratio ? Ans. Theorem 4th gives r=2. 3. Find the sum of the geometrical series, 1, 5, 4, 5, &c., continued to infinity. Ans. 2. 4. Find the sum of the geometrical series, whose first term is 1, and common ratio , when continued to infinity. Ans. 5. 5. Find the sum of the geometrical series whose first n-1 term is m and common ratio when continued to infinity Ans. mn. 6. A servant agreed with his master to serve him for twelve months, upon this condition, that for his first month's service he should receive a farthing, for the second a penny, for the third fourpence, and so on. What did his wages amount to at the expiration of his service ? Ans. L.5825, 8s. 5 d. 7. There are three numbers in geometrical progression whose sum is 52, and the sum of the first and second is to the sum of the first and third as 2 is to 5. Required the numbers. Ans. 4, 12, 36. 8. There are three numbers in geometrical progression. The sum of the first and second is 15, and the sum of the first and last is 25. What are the numbers ? Ans. 5, 10, 20. GEOMETRICAL RATIO. 97. The geometrical ratio between two numbers is determined by dividing the one number by the other. The quotient is the value of the ratio. The number divided is called the antecedent, and the divisor the consequent of the ratio. Thus the ratio of 9 to 6 is g=lį, in which 9 is the antecedent and 6 the consequent, and the value of the ratio is 1, Ratio may therefore be considered as a fraction, the numerator of which is the antecedent, and the denominator the consequent of the ratio. When the antecedent is greater than the consequent, it is called a ratio of greater inequality, and when the antecedent is less than the consequent, it is called a ratio of lesser inequality. 98. Since ratios can be expressed by fractions, they can |