We already know the value of S in terms of a, n, and r. a=l-(n-1)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 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 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 1, and m+2 for n, it becomes that is, the 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 b-a to the first term a. The second mean is obtained by augm+1' menting the first by r, &c. 1. Find 3 arithmetical means between the extremes 2 and 18. The formula 2. Find 12 arithmetical means between 12 and 77. The formula 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.c.de.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 which are equal to each other, since a, b, c are in progres sion: therefore, the common difference is the same in each of the first term of the second, &c, we may conclude that all of these partial progressions form a single progression. partial progressions; and since the last term of the first, forms the EXAMPLES. 1. Find the sum of the first fifty terms of the progression 2.9.16.23 For the 50th term we have l=2+49×7=345. Hence, 50 S=(2+345)×2=347×25=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 and the number of terms 21: required the sum of the series. Ans. 140. 1 3' 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 nth, which contains n. In other words, find the expression for the sum of the natural numbers 1, 2, 3 from 1 to n, Ans. S=n(n+1) inclusively. ...و 2 9. Find the sum of the n first terms of the progression of uneven numbers 1, 3, 5, 7, 9 Ans. Sn2. 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 ? 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 BD 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. To express that the ratio of A to B is equal to the ratio of C to D, we write the quantities thus, A:B::C: D. and read, A is to B, as C to D. The quantities which are compared together are called the terms of the proportion. The first and last terms are called the two extremes, and the second and third terms the two means. 170. Of four proportional quantities, the first and third are called the antecedents, and the second and fourth the consequents; and the last is said to be a fourth proportional to the other three taken in order. 171. Three quantities are in proportion when the first has the same ratio to the second that the second has to the third; and then the middle term is said to be a mean proportional between the other two. 172. Quantities are said to be in proportion by inversion, or inversely, when the consequents are made the antecedents and the antecedents the consequents. 173. Quantities are said to be in proportion by alternation, or alternately, when antecedent is compared with antecedent and consequent with consequent. 174. Quantities are said to be in proportion by composition, when the sum of the antecedent and consequent is compared either with antecedent or consequent. 175. Quantities are said to be in proportion by division, when the difference of the antecedent and consequent is compared either with antecedent or consequent. 176. Equi-multiples of two or more quantities are the products which arise from multiplying the quantities by the same number. |