Solving this quadratic and taking the positive value of the root, we find, x = 30 N.B. The length of the whole journey, 120 miles, does not enter into the solution. It will now merely serve to verify the result, and to indicate that the first fast train travelled 5 miles before it came up with the luggage train. SECTION II. 1. "Prove the formula for finding the sum of an arithmetical series. (a) "The first term of an arithmetical progression is 1; the common difference, 1; the sum of the series, 36. Required the number of terms." (b) (a) Let a first term and d = common difference. Now the 2nd term 1st term + d = a + d * 3rd term = 2nd term + d = a + 2d 4th term 3rd term + d = a + 3d * * * nth term = (n-1)th term + d = a + n~ 1. d Now S, the sum of the series, is equal to all these terms put together; or, S = a + (a + d) + (a + 2d) + (a + 3 d) +.. +(a + n−1. d) .... This series may be written in an inverse order; thus, S = (a + n −1. d) + .... + (a + 3d) + (a + 2d) +(a + d) + a. And now the addition of any two terms that are vertically opposite in the series, will give (2a+n-1.d). For the sum of the first terms in each line this is sufficiently evident; the second term in the second line is d less than the preceding term; that is, it is (a + n-2. d). This added to (a + d) the second term of the first line = (2a + n— -1. d). The third term of the second line is (a + n -3. d); the third term of the first line is (a + 2d); the sum of these is (2 a + n−1. d), as before. The quantity (2 a+n-1. d) is repeated as many times as there are terms in the progression, that is n times in the present case. Hence, by adding the corresponding term of each line, we get 2S (2a+n—1. d) + (2 a + n−1. d)+ &c. to n terms 2S = ..S=(2a+n—1. d) 1⁄2 ........ (1) 2 =(a+1); where l is the last term of the series, 2 and is equal to the first term, a, plus the common difference multiplied by the number of its place in the series less by one. (b) In the example given S = 36; a = 1; d 1; and n is to be found. Since 7 is not given, formula (1) 2 (a) "When are quantities said to be proportional? (b) and when is one quantity said to vary as another? (c) Prove that if a b c d; then a : a ~ b C~ d." c: (a) When the ratio between two quantities a, b, is the same as between two other quantities c, d; or when abc÷d the four quantities form a proportion which is written a b c d, in which a is the same multiple, part, or parts of b, that c is of d.(b). One quantity x, varies as another quantity y, when the magnitude of x depends on that of y. Thus, if x be the length and y the breadth of a parallelogram, and it be required to construct a similar parallelogram, having a given ratio to the first, x varies as y, or the length and breadth increase or diminish simultaneously. If a parallelogram be required equal in area but of different length from another parallelogram; then, as the length increases, the breadth diminishes, and the converse. In this case x varies inversely as y. that is, a b : a :: c ±d: c or, inversely, a: a ±b::c: c±d....(1) Whence ba: a : d±c:c And inversely ab±a: cdc....(2) But a: a b c: c±d 3. "Find the number of different combinations that may be made of n different things taking r of them together." Of n different objects, a, b, c, d, &c., a may be placed before each of the rest and thus form n-1 permutations in which a stands first; there may similarly be n― -1 permutations in which b stands first; and the same for each of the n different objects; so that there are n (n-1) permutations of objects taken two and two. (1) If from then things the first be taken away there will remain n-1 things; and the number of permutations that can be formed of them, taken two and two, will he (n-1) (n-2). Now let the object a which we have assumed to be taken from the n things be prefixed to each of the permutations (n-1) (n-2), and there will be (n-1) (n-2) such permutations taken three together, in which a stands first; there will be as many in which b stands first, and so on for the rest; so that, on the whole, n (n−1) (n—2) = number of permutations of n different things taken three together. We thus perceive that the number of permutations o n things taken -1) two together n N Similarly it is shown, * four together n (n-1) -2) n- -3) five together.. n (n-1) (n—2) (n—3) (n—4) r together .... n (n-1) (n-2) (n−3) (n−4) (n−5) =n (n-1) (n-2) (n-3)....(n—r + 1) For each combination of things taken two and two there are twice as many permutations; for each combination, as a b, admits of two permutations b a, a b. Hence the permutations of things taken two and two must be divided by two to give the combinations of the same objects; or the number of combinations of n n (n-1) things taken two and two together 2 There are n (n-1) (n-2) permutations of n things taken three and three together, and each combination of three admits 3 x 2 x 1 permutations, so that the permutations must be divided by 3 x 2 x 1 to give the combinations; that is, the number of combinations of n things taken three and three, is taken four together n (n-1) (n-2) = 1.2.3 n (n-1) (n-2) (n—3) 1.2.3.4. And similarly the number of combinations of n things taken r times together n (n-1) (n-2) (n−3) .... (n—r+1) 1 2 3.4 r Example; required the number of parties of three that may be formed out of six persons. 1. "In a circle the angle in a semicircle is a right |