EXAMPLE VIII.-Insert 4 arithmetical means between 37 and 7. The number of means being 4, the number of terms is (4+2)6. The common difference, therefore (§ 252), is 37-7 30 = :6; so that (249) the required means are 6-I 5 (37-6-) 31, (31-6=)25, (25-6-)19, and (19-6-)13. The progression in full is 37 31 25 19 13 7 EXAMPLE IX.-Insert one arithmetical mean between 13 and 27. - =7; so that There being only one mean, the number of terms is (1+2=)3. 27 13 14 The common difference, therefore, isthe required mean is (13+7=) 20. 3-I 2 EXAMPLE X.-Insert one arithmetical mean between 23 and 15. As in the last case, the number of terms is (1+2=)3; so that the common difference is 23-15_8_ =4. The required mean, therefore, is (23-4)19. 3-I 2 ONE arithmetical mean, however, can be inserted more easily. 254. To insert ONE arithmetical mean between two numbers: Add the numbers together, and take half their sum. Thus, the arithmetical mean between 13 and 27 (Ex. IX.) is 13+27 40 =20; and the arithmetical mean between 23 and In order to understand this, let us put a for the arithmetical mean between 13 and 27, and suppose that a boy has 13 marbles in his left-hand pocket, and 27 in his right-hand pocket. Now, as 13, x, and 27 form an ascending arithmetical progression (x being greater than 13 by as much as 27 is greater than x), it is evident that, by transferring from his right-hand to his left-hand pocket a number of marbles equal to the common difference, the boy would have x marbles in each pocket. thus find x+x=13+27; 2x=13+27; x=' 13+27 2 We Next, let us put y for the arithmetical mean between 23 and 15, and suppose that a boy has 23 marbles in his left-hand pocket, and 15 in his right-hand pocket. As 23, y, and 15 form a descending arithmetical progression (23 exceeding y by as much as y exceeds 15), it is obvious that, by transferring from his left-hand to his right-hand pocket a number of marbles equal to the common difference, the boy would have y marbles in each pocket. Consequently, y+y=23+15; 2y= 23+15; y=23+15. 2 255. The sum of the extremes of an arithmetical progression is the same as the sum of any two means equally distant from them-the same, for instance, as the sum of the first mean and the last; the same as the sum of the first mean but one and the last but one; &c. As an illustration, let us take the ascending progressionI 3 5 7 9 I I 13 15 17 19 21 23 If we suppose that a boy has I marble in his left-hand pocket, and 23 in his right-hand pocket, we shall see that the transfer of 2 marbles (2 being the common difference) from the righthand to the left-hand pocket would leave 21 marbles in the right-hand, and 3 in the left-hand pocket; that the transfer of two more in the same direction would leave 19 in the righthand, and 5 in the left-hand pocket; that another such transfer would leave 17 in the right-hand, and 7 in the left-hand pocket; that another such transfer would leave the contents of the pockets 15 and 9, respectively; and another, 13 and 11, respectively. Hence, 1+23=3+21=5+19=7+17=9+15= 11+13. As a second illustration, let us take the descending arithmetical progression 53 48 43 38 33 28 23 18 13 8 3 Supposing that a boy has 53 marbles in his left-hand, and 3 in his right-hand pocket, we see that, by transferring 5 (the common difference) a number of times from the left-hand to the right-hand pocket, he would have first, 48 in the left-hand, and 8 in the right-hand pocket; next, 43 in the left-hand, and 13 in the right-hand pocket; then, 38 in the left-hand, and 18 in the right-hand pocket; then, 33 in the left-hand, and 23 in the right-hand pocket; and, lastly, 28 in each pocket. Hence, 53+3=48+8=43+13=38+18=33+ 23-2828. From this last illustration it will be seen that the middle term of an arithmetical progression which contains an odd number of terms is half the sum of the extremes; and that the middle one of any three consecutive terms is half the sum of the other two. 256. To find the sum of the terms of an arithmetical progression: Multiply the sum of the extremes by half the number of terms. This follows from § 255. The sum of any two terms equally distant from the extremes being the same as that of the extremes themselves, all the terms, taken together, contain as many such sums as there are pairs of terms, and contain, in addition, the half of such a sum when the number of terms is odd-the middle term being, as we have seen, half the sum of the extremes. Thus, the sum of the terms of the progression I 3 5 7 9 I I 13 15 17 19 21 23 can be found when the extremes and the number of terms are known. For, from what has already been explained, it is evident that, the sum of the extremes being (1+23=)24, all the terms, taken together, contain as many twenty-fours as there are pairs of terms; so that, the number of terms being 12, and the number of pairs 6, the sum of all the terms is 24 × 6=144. Let us next take the progression 53 48 43 38 33 28 23 18 13 8 3 The sum of the extremes being (53+3=)56, and the number of terms II, the sum of all the terms is 56x=56×51= 308. Because the middle term (28) is half the sum of the extremes; and the remaining terms, taken together,—being 10 in number, contain 5 such sums. 257. To find the number of terms in an arithmetical progression-the sum of the extremes and the sum of all the terms being given: Divide the sum of all the terms by the sum of the extremes, and double the quotient. This follows from § 256. The sum of all the terms being the product of the sum of the extremes by half the number of terms, we obtain half the number of terms on dividing the sum of all the terms by the sum of the extremes. Thus, if the sum of all the terms were 3502, and the sum of the extremes 412, half the number of terms would be (3502÷412=) 8), and the number of terms (81×2=) 17. 258. To find either extreme of an arithmetical progression-the other extreme, the number of terms, and the sum of the terms being given: Divide the sum of the terms by half the number of terms, and subtract the given extreme from the quotient. This, also, follows from § 256. The division of the sum of the terms by half the number of terms gives the sum of the extremes for quotient, which, therefore, exceeds the given extreme by the required extreme. Thus, if the sum of the terms were 1,886, the number of terms 23, and one of the extremes 5, the sum of the extremes would be (1,886÷23=)164, and the required extreme (164—5=) 159. 259. The sum of the terms of the progression I 3 5 7 9 &c. is the square of the number denoting how many terms there are. Thus, the sum of the first two terms (1+3) is the square of 2; the sum of the first three terms (1+3+5), the square of 3; the sum of the first four terms (1+3+5+7), the square of 4; and so on.* GEOMETRICAL PROGRESSION. 260. Every term (after the first) of a geometrical progression is the product of the preceding term by the common ratio-the progression being an ascending" or a "descending" one according as the common ratio is greater or less than unity. EXAMPLE I.-Set down the geometrical progression which has 5 for its first term, and 2 for common ratio. Multiplying 5 by 2, we obtain the second term, 10; multiplying 10 by 2, we obtain the third term, 20; multiplying 20 by 2, we obtain the fourth term, 40; and so on. The pro gression, therefore, is— 5 ΙΟ 20 40 80 160 320 640 1280 2560 &c. EXAMPLE II.-Set down the geometrical progression which has 81 for its first term, and for common ratio. * Putting n for the number of terms, and remembering that the common difference is 2, we have I + (n-1) x 2 the last term (§ 250); so that (§ 256) the sum of n terms is 1+1+(n−1) × 2 x== x2} Multiplying 81 by, we obtain the second term, 27; multiplying 27 by, we obtain the third term, 9; multiplying 9 by , we obtain the fourth term, 3; and so on. The progression, therefore, is 1 81 &c. 81 27 9 3 I 1/8 13 27 BI The preceding progressions-of which the first is an ascending, and the second a descending one-might be written as follows: 5 5×2 5x22 5x23 5X24 5X25 5x26 5X27 &c. 81 81× 81x()2 81x(3) 81x(3) 81x() &c. 4 5 So that every term (after the first) of a geometrical progression is the product of the first term by a power of the common ratio, the index of which power is less by I than the number denoting the place of the term.* 261. The first term and the common ratio being given, to find any other term of a geometrical progression: Subtract one from the number denoting the place of the required term; raise the common ratio to the power indicated by the remainder; and multiply the result by the first term. EXAMPLE III.-The first term of a geometrical progression is II, and the common ratio 5; find the seventh term. Seventh term first term x sixth power of common_ratio= 11 × 56=11X15,625=171,875. EXAMPLE IV.-The first term of a geometrical progression is 256, and the common ratio ; find the tenth term. Tenth term first term X ninth power of common_ratio= 256X() 256x19683=19224 262. The extremes and the common ratio being given, to find the number of terms in a geometrical progression: Divide the first extreme by the last; find what power of the common ratio the quotient is; and add 1 to the index of this power. EXAMPLE V.-The extremes of a geometrical progression are 234,375 and 3, and the common ratio is 5; find the number of terms. * Putting ƒ for the first term, and for the common ratio, we have2nd term=fXr 3rd 4th nth |