the second, &c., travelling each day 2 miles less than the preceding. The second goes 21 miles the first day, 24 the second, &c., increasing each day 3 miles. What will be the distance between them at the end of ten days? Ans. 15 miles. PROBLEMS IN ARITHMETICAL PROGRESSION, PRODUCING EQUATIONS. (92.) 1. There are three numbers in arithmetical progression, their sum is 18, and the sum of their squares 158. What are those numbers? Let = the first, and y x y the last, then + will be the other (81), and we shall have the equations, Clearing the first equation of fractions, we obtain, 2 Putting this value of (x+y) in the second equation, that equation becomes, From the equation x+y= 12, we have, y= 12-x. Squaring this value of y, and putting it in the place of y2, in the last equation, we have, Transposing and reducing, x2- 12x — — 11, x2-12x+36-11+36= 25, 6. The numbers are therefore 1, 6, and 11. Another Solution. Let x the second number, and y = the common differ = ence; then x y, x, and xy will be the numbers (79), and we shall have the equation, and reducing, x−y + x + x + y = 18; 3x= 18, x= 6. By the second condition, the sum of their squares is 158; therefore, (6 − y)2 +(6)2 + 6 + y)2 = 158 ; or, 36-12y + y2+36 +36 + 12y+y2 = 158. Transposing and reducing, 2y2 = 50, y2 = 25, x—y = 6—5 = 1, x = 6, x+y=6+5=11. 1, 6, and 11, are therefore the numbers. 2. There are five numbers in arithmetical progression; the sum of these numbers is 65, and the sum of their squares 1005. What are the numbers? Let x = the first, and y = shall have (79), = the common difference, then we x+(x+y)+(x+2y) + (x + 3y) + (x+4y) = 65 x2 +(x+y)2 +(x+2y)2 + (x + 3y)2 + (x + 4y)2 = 1005, By performing the operations indicated, and reducing, these equations become x + 2y = 13, x2+4xy + 6y2 = 201. Squaring the first of these equations, and subtracting it from the second, there remains 2y = 32; from which we find y = 4, and by putting this value of y in the first equation, we find X = 5. The numbers are therefore 5, 5+4, 5+8, 5+12, 5+16, or, 5, 9, 13, 17, and 21. In the same manner we might find any number of terms in an arithmetical progression, their sum, and the sum of their squares being given. 3. A number consists of three digits which are in arithmetical progression; and if the number be divided by the sum of its digits, the quotient will be 26; but if 198 be added to it, the digits will be inverted. Required the number. Ans. 234. 4. A person bought 7 books, the prices of which were in arithmetical progression. The price of the next above the cheapest was eight shillings; and the price of the dearest 23 shillings. What was the price of each book? Ans. 5, 8, 11, 14, 17, 20, and 23 shillings, respectively. 5. A and B, 165 miles distant from each other, set out with a design to meet; A travels 1 mile the first day, 2 the second, 3 the third, and so on; B travels 20 miles the first day, 18 the second, 16 the third, &c. In how many days will they meet, and how far will each have travelled? Ans. They meet in 10 days, A having travelled 55 miles, and B 110. GEOMETRICAL PROGRESSION. (93.) A geometrical progression, or a progression by quotients, is a series of numbers, having this property, namely, each term of the series, except the first, is the product of the one that precedes it by a constant number. This number is called the ratio of the progression. A geometrical progression is written in the following manner (Arith. 72):-26 18:54 : 162: 486. : In this progression the ratio is 3, each term being three times that which precedes it. When the ratio is greater than unity, the progression is called an increasing progression; and when the ratio is less than unity, it is called a decreasing progression; 812793: 1 : 1 : 1 : 2 is a decreasing progression, the ratio of which is . In order to investigate the properties of geometrical progression, let the first term be represented by a, the last by 7, the ratio by r, the sum of all the terms by S, and the number of terms by n. (94.) Since each term of the progression is produced by multiplying the term that precedes it by the ratio, the first term being a, the second will be a xr, or ar; the third, ar ×r, or ar2; the fourth, ar3, and so on. We shall have, therefore, for the general expression of a geometrical progression, In this progression, it will be perceived, that the exponent of r in each term, is equal to the number of terms that precede it; and hence, in an equation of n terms, the last term having 1 terms before it, will be ar"-1; and calling this last term l, as was proposed, we have, n 1 = ar"-1. Hence we conclude that, in every geometrical progression, the last term is equal to the first term, multiplied by that power of the ratio, which is equal to the number of terms less 1 (Arith. 74). (95.) Let x be a term of a geometrical progression which P terms before it, we shall have (94), has x= arp. Let y be a term of the same progression which has p terms after it; then, since the exponents of r diminish by unity, for each term, proceeding from the last towards the first, the term which has p terms after it, will be represented by ar"--"; wherefore, y: = ar N--1--P Multiplying this equation, term by term, by the preceding, there results, xy=a22+” = a2p”--1. If the progression have an odd number of terms, and a be the middle term, then a y, and we shall have, But in any geometrical progression, a being the first term, and ar"- the last, we shall have, a×ar", or a2r2--1, for the product of the extremes. Hence we conclude that, in every geometrical progression the product of the extreme terms is equal to the product of any two mean terms, equally distant from them; and also equal to the square of the middle term, when there is an odd number of terms (Arith. 74). (96.) Let a b c d : e, &c. be a geometrical progression. From the definition of a geometrical progression, we shall have, b ar, c = = br, d = cr, e = dr.. = Adding these equations together, we obtain, = kr. b+c+d+e....+i ar+br+cr + dr+er....+kr, = or, b+c+d+e....+1 = (a+b+c+d+e....+k)r. The first member of this equation contains all the terms of the progression except a; and hence if S be the sum of all the terms, it is equal to S— a. The second member contains all the terms of the same progression except 7, mulplied by r; it is therefore equal to (S' — l)r ; whence, S-a (S-1)r; or, S-arS-rl. = rl a Transposing, rS—S=rl — a ; whence, S= r— 1° From which it appears, that the sum of all the terms of a geometrical progression, is equal to the product of the last term by the ratio, minus the first term, divided by the ratio diminished by unity (Arith. 75). (97.) The equations = ar"-1, and S = rl. a , express the 1 r relations which exist between the quantities a, n, l, r, and S, and furnish the means of determining any two of these quantities, the other two being known. If in the equation S= rl-a 7 1' we substitute in the place of l its value ar1, as found in the other, this last equation becomes By means of this formula, the sum of a progression may be determined without knowing the last term. rl- a (98.) The equation S = gives, by clearing of frac tions and reducing, rS-Srl - a, or r$ — rl = S— a; From which we learn, that the ratio of any progression is equal to the sum, minus the first term, divided by the sum, minus the last term (Arith. 76). (99.) If in the equation S= 1) we consider r great 1 er than unity, it is evident that S may be made greater than any quantity however large; for the quantity r" may be made greater than any assignable quantity by increasing |