This being proved, write the progression below itself, but in an inverse order, viz., . a.b.c.d.e.f. . . . i . k .1. Calling <S the sum of the terms of the first progression, 2S will be the sum of the terms in both progressions, and we shall have 2S=(a+Z)+(6+A)+(c-H) . . . +(i+e)+(k+b) +(l+a). Now, since all the parts a+l, b+k, c+» .... are equal to each other, and their number equal to n, 2S={a+l)n, or S=(^—^n. That is, the sum of the terms of an arithmetical progression is equal to half the sum of the two extremes multiplied by the number of terms. 1. The extremes are 2 and 16, and the number of terms 8: "what is the sum of the series? S=(£±!)x», gives S=^x8=72. 2. The extremes are 3 and 27, and the number of terms 12: what is the sum of the series 1 Ans. 180. 3. The extremes are 4 and 20, and the number of terms 10: what is the sum of the series? Ans. 120. 165. The formulas l=a+(n-l)r and S=^xn contain five quantities, a, r, n, I, and S, and consequently give rise to the following general problem, viz.: Any three of these five quantities being given, to determine the other two. I We already know the value of S in terms of a, n, and r. a=l—(n—l)r. That is, the first term of an increasing arithmetical progression is equal to the last term, minus the product of the common difference by the number of terms less one. , From the same formula, we also find l-a That is, in any arithmetical progression, the common difference is equal to the difference between the two extremes divided by the number of terms less one. 1. The last term is 16, the first term 4, and the number of terms 5: what is the common difference? The formula I—a . 16 r= mves r= n— 1 b 2. The last term is 22, the first term 4, and the number of terms 10: what is the common difference? Ans. 2. 166. The last principle affords a solution to the following question: To find a number m of arithmetical means between two given numbers a and b. To resolve this question, it is first necessary to find the common difference. Now we may regard a as the first term of an arithmetical progression, b as the last term, and the required means as intermediate terms. The number of terms of this progression will be expressed by m+2. Now, by substituting in the above formula, b for I, and m+2 for n, it becomes that is, <Ae common difference of the required progression is obtained by dividing the difference between the given numbers a and b, by one more than the required number of means. Having obtained the common difference, form the second term of the progression, or the first arithmetical mean, by adding r, or . , °, to the first term a. The second mean is obtained by augm+1 '6 raenting the first by r, &c. 1. Find 3 arithmetical means between the extremes 2 and 18. The formula b-a 18-2 A r=^+I glves r=~T~=4i hence, the progression is 2 . 6 . 10 . 14 . 18. 2. Find 12 arithmetical means between 12 and 77. Th« formula b-a 77—12 . r=z——r gives r=———=5. Hence the progression is 12 . 17 . 22 . 27 72 . 77. 167. Remark. If the same number of arithmetical means are inserted between all of the terms, taken two and two, these terms, and the arithmetical means united, will form but one and the same progression. For, let a . b . e . d . e .f . . . . be the proposed progression, and m the number of means to be inserted between a and b, b and c, c and d . . . . From what has just been said, the common difference of each partial progression will be expressed by b—a c—b d—e which are equal to each other, since a, b, c . . . are in progression: therefore, the common difference is the same in each of the partial progressions; and since the last term of the first, forms the first term of the second, &c, we may conclude that all of these partial progressions form a single progression. EXAMPLES. 1. Find the sum of the first fifty terms of the progression 2 . 9 . 16 . 23 . . . For the 50th term we have 2=2 + 49x7=345. 50 Hence, S=(2+345) x —=347X25=8675. 2. Find the 100th term of the series 2 . 9 . 16 . 23 . . . Ans. 695. 3. Find the sum of 100 terms of the series 1.3.5.7.9... Ans. 10000. 4. The greatest term is 70, the common difference 3, and the number of terms 21: what is the least term and the sum of the series? Ans. Least term 10; sum of series 840. 5. The first term of a decreasing arithmetical progression is 10, the common difference -i-, and the number of terms 21: required a the sum of the series. Ans. 140. 6. In a progression by differences, having given the common difference 6, the last term 185, and the sum of the terms 2945: find the first term, and the number of terms. Ans. First term =5; number of terms 31. 7. Find 9 arithmetical means between each antecedent and consequent of the progression 2.5.8.11.14 . . . Ans. Ratio, or r~0,3. 8. Find the number of men contained in a triangular battalion, the first rank containing 1 man, the second 2, the third 3, and so on to the n"1, which contains n. In other words, find the expression for the sum of the natural numbers 1,2,3 . . ., from 1 to n, inclusively. Ans. 9. Find the sum of the n first terms of the progression of uneven numbers 1, 3, 5, 7, 9 . . . Ans. S=n2. 10. One hundred stones being placed on the ground, in a straight line, at the distance of 2 yards from each other, how far will a person travel, who shall bring them one by one to a basket, placed at 2 yards from the first stone 1 Ans. 11 miles, 840 yards. Geometrical Proportion and Progression. 168. Ratio is the quotient arising from dividing one quantity by another quantity of the same kind. Thus, if A and B represent quantities of the same kind, the ratio of A to B is expressed by B 169. If there be four magnitudes, A, B, C, and D, having such values that B ID then A is said to have the same ratio to B, that C has to D; or, the ratio of A to B is equal to the ratio of C to D. When four quantities have this relation to each other, they are said to be in proportion. Hence, proportion is an equality of ratios. |