EXAMPLES. (1) Find the nth term and the sum of n terms of the series of odd numbers, 1, 3, 5, 7, &c. n Generally, a + (n-1) d, and s= {2a + (n−1)d}; 2 here a 1, and d = 2; whence, by substitution, (2) Find the nth term and the sum of n terms of the series of even numbers, 2, 4, 6, 8, &c. n and s= {4 + (n − 1) 2} = n (n + 1). 2 (3) Required the 7th term, and the sum of 7 terms of the series 11, 8, 5, &c. Here a= 11, n = 7, and d - 3 (4) If the first term of an arithmetical progression be 2, and the sum of 8 terms 100, what is the common difference? In the case proposed, s = 100, a = 2, and n = 8; Hence, the series is 2, 5, 8, 11, 15, 18, 21, 24. (5) If the sum of 9 terms of an arithmetical progression be 171, and the common difference 4, what is the first term? and substituting 171 for s, 9 for n, and 4 for d, Whence, the series is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39. EXAMPLES FOR PRACTICE. (1) Find the sum of 12 terms of the series 8, 15, 22, &c. Ans. 558. Ans. (2) Required the 7th term, and the sum of 7 terms of the series,,, &c. , and 0. (3) Required the number of terms and the common difference, when the sum is 2.748, the first term .034, and the last term .0576. Ans. 60, and .0004. (4) What number of terms of the series 54, 51, 48, &c. must be taken to make their sum 513? Ans. 18 or 19. (5) The number of deaths in a besieged garrison amounted to 6 daily; and, allowing for this diminution, their stock of provisions was sufficient to last 8 days. But, on the evening of the 6th day, 100 men were killed in a sally, and afterwards the mortality increased to 10 daily. Supposing the Р stock of provisions unconsumed at the end of the 6th day sufficient to support 6 men for 61 days; it is required to find how long it would support the garrison, and the number of men alive when the provisions were exhausted. Ans. 6 days, and 26 men. 161. To insert any number of arithmetical means between any two given quantities. Let a and be the two given quantities, and m the number of means required to be inserted between them; then, since the number of terms = m +2, we shall have Ex. Insert three arithmetic means between 117 and 477. whence the series is 117, 207, 297, 387, 477. The same might be obtained by substituting for a, m, and 7, their values, in the general series. EXAMPLES FOR PRACTICE. (1) Insert 4 arithmetic means between 193 and 443. Ans. The means are 243, 293, 343, 393. (2) Find 2 arithmetic means between 3 and 3. (3) Insert 9 arithmetic means between 1 and Ans.,,,, 0, −1, −1, −3, -3. 162. Any two terms and their places being given, to determine the series. Let P and Q represent the pth and 9th terms respectively, and the rest as before; then, P= P = a + (p − 1)d, and Q = a + (g − 1) d ; whence PQ = (p − q) d, and .. d = .. a = P(p − 1) d = P – (p-1) 1 Р p-q Hence, the first term and common difference being found, the series may be determined. Moreover, the nth term = a + (n − 1) d GEOMETRICAL PROGRESSION. 163. A Geometrical Progression is a series of quantities in continued proportion, which therefore increase or decrease by a common ratio. If a be the first term and r the common ratio, the series will be a, ar, ar2, ar3, &c. 164. Hence, if the series be given, the common ratio may be found by dividing the second term by the first, or any other term by that which immediately precedes it. 165. If quantities be in geometrical progression, their differences are in geometrical progression. Leta, ar, ar2, ar3, &c. be the quantities: their differences ara, ar? ar, ar3 — ar2, &c. form a geometrical progression, whose first term is ar a, and common ratio r. 166. Quantities in geometrical progression are proportional to their differences. For, a, ar, ar2, ar3, &c, a: ar :: ur—a : ar2. —ar :: being the quantities, ar2 — ar : ar3 — ar2 &c. 167. In any geometrical progression, the first term is to the third, as the square of the first to the square of the second. Let a, ar, ar2, ar3, &c. be the progression; then a: ar2 :: a2 : a2r2. In the same manner it may be shewn, that the first term is to the (n+1)th, as the first raised to the nth power, to the second raised to the same power. 168. The terms taken at equal intervals, in a geometrical progression, are in geometrical progression. For, of the progression a, ar, ar2, ... arn, ... ... &c. the terms a, ar", ar2n, &c., taken at the interval of n terms, form a geometrical progression, whose common ratio is r”. |