ARITHMETICAL PROGRESSIONS, OR EQUIDIFFERENT SERIES. 352. We have shown that when a series of numbers increase or decrease by the same quantity, those numbers are said to be in arithmetical progression. Thus 3, 6, 9 12, 15, 18, &c., or 21, 17, 13, 9, 5. 1,—3,—7, &c. It must be observed that any term is composed of the preceding term, plus or minus the common difference. For instance, 6 consists of the first term 3 and of the common difference 3; 9 consists of the second term 6 and of the common difference 3; similarly, 17 consists of the first term 21, minus the common difference 4; 13 consists of 17 minus 4, &c. 353. The natural series of numbers, 1, 2, 3, 4, 5, 6, &c., determine an arithmetical progression, the first term of which is 1, and the common difference 1. The even numbers 2, 4, 6, 8, 10, &c., as well as the uneven numbers 1, 3, 5, 7, 9, &c., constitute an arithmetical progression, the common difference being 2. The series of numbers, 10, 20, 30, 40, &c., form an arithmetical progression, having 10 for its common difference. 354. From what has been mentioned on the formation of the terms of an arithmetical progression, we are enabled to determine any term of such a series. For an increasing arithmetical progression we have: 3rd 4th 5th 6th =3+3+3, or 3+2×3, or 9. =3+3+3+3, or 3+3 × 3, or 12. =3+3+3+3+3+3, or 3+5 × 3, or 18. 7th =3+3+3+3+3+3+3, or 3+6 × 3, or 21. Likewise, for a decreasing series, we have: The 1st term=21. =21-4, or 21-1 × 4, or 17. =21-4 =21-4 =21-4 -4-4-4, or 21-6 × 4, or-3. It follows, then, that any term consists of the first plus or minus (according as the equidifference is increasing or decreasing) the common difference, multiplied by the number of terms preceding the one required. The seventh term is composed of the first plus, or minus six times the common difference. 355. Let us represent the first term of an A.P. by a, the common difference by d, then the increasing series is : 1st, 2nd, 3rd, 4th. 5th, 6th, 7th nth term. a, a+d, a+2d, a+3d, a+4d, a +5d, a+bd...a+(n-1)d. And the decreasing series is : 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, nth term. and a, a―d, a—2d, a—3d, a—4d, a—5d, a—6d...a—(n—1)d. Which increasing and decreasing series may be thus written: 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, nth term. a, a±d, a±2d, a±3d, a+4d, a+5d, a±6d...a±(n−1)d. Where it is seen that any term is found by adding to or subtracting from the first term, the common difference, multiplied by the number of terms which denote the place in the series, minus 1. 356 If stands for the last or nth term of a series, then l=a+(n-1)d. Ex. Let the first term of an increasing series be 4, and the common difference 5, then the 24th term of the A.P.=4+ 23×5=119. The 50th term 4+49x5=249. Or, if in a decreasing A.P. the first term be 248, and the common difference 2, then the 13th term=248-12 × 2=224. The hundredth term = 248-99 × 2=50. 357. In the expression l=a+(n-1)d, which represents the nth term of an A.P., let us subtract a from each side of the sign of equality of the increasing series, then : l=a+(n-1)d. Hence the common difference is equal to the last term, minus the first, divided by the number of terms, minus 1. This is evident, since the last term consists of the sum of the first term, and a certain number of times the common difference; l-a, expresses that certain number of times the common difference, and this remainder being divided by the number the common difference is multiplied by, the quotient is necessarily the common difference. Again, if in a decreasing progression we have la-(n—1) d, adding to each side of the sign of equality (n-i) d, then (n-1) d +1=a, from which I being subtracted from these equal sums, we have (n-1) da-l, and dividing these equal remainders by a-l (n-1), it is found that d= n-l Here the common difference is equal to the first term minus the last, divided by the number of terms minus 1. Example 1. Suppose an A.P. of 14 terms, in which the first term a=3, and the last 155, it is required to find the common difference d. and the A.P. is 3, 7, 11, 15, 19.........55. Ex. 2. What is the common difference of a series consisting of 10 terms, the first term is 11 and the last 4? and the equi-difference is 11, 103, 43, 88, 73.......4. 358. In a similar manner do we determine n, the number of terms, by means of a, the first term, 7 the last, and d the common difference. As before, la+(n-1) d. Let a be subtracted from each side of the sign of equality, then la(n-1) d, dividing both equal remainders by d, it follows that =n-1, and adding 1 to both equal quantities, we have 1+ l-a -a Hence the number of terms is found by dividing the last term, minus the first, by the common difference and adding 1 to the quotient. For taking away the first term from the last, the remainder is the common difference multiplied by a certain number, which being divided by the common difference, the quotient is the number by which the common difference was multiplied; this number is 1 less than the number of terms, therefore 1 must be added to the quotient. For a decreasing A.P. we have: l-a-(n-1) d, adding to both sides of the sign of equality (n-1) d. .. l+(n−1) d=a, substracting from both sides of the sign of equality, we have (n-1) d=a―l, which being divided by d, gives N -1 and adding to both sides 1, d Hence the number of terms is found, &c. 359. If we were to subtract (n-1) d from both sides of l=a +(n-1) d, we have l-(n-1) d=a. Therefore the first term of an A.P.=the last term, minus the product, of the common difference and the number of terms less 1. This truth is rendered evident when we consider that the last term consists of two parts, the first term and the product of the common difference and the number of terms minus 1; then subtracting the second part, the remainder is the first term. When the A.P. is decreasing, we have: l—a—(n—1) d, and adding (n—1) d to each side, l+(n−1) d = a. Hence the first term is found, &c. Example 1. The last term is 53, the common difference 3, and the number of terms 18 Required the first term. a=53—(18—1)3=2. Ex. 2. The last term of an A.P. is 5, the common difference 2, and the number of terms 52. What is the first term? a=5+(52-1)2=107. 360. An important case in A.P. is to find the sum of the terms; the operation of adding all the successive terms together would be very tedious; we shall now determine a method by means of which the sum of the terms of any A.P. can at once be found. Let it be required to ascertain the sum of the terms of the following A.P. 3, 7, 11, 15, 19, 23, 27, 31, 35, 39. In order to determine the sum of the terms, let us write the progression, both as an increasing and decreasing series, setting the terms in order, and adding together the corresponding ones, as follows: Increasing series... 3, 7, 11, 15, 19, 23, 27, 31, 35, 39. Decreasing series...39, 35, 31, 27, 23, 19, 15, 11, 7, 3. 42, 42, 42, 42, 42, 42, 42, 42, 42, 42. We obtain every term 42, as the sum of two terms, and this sum is repeated as many times as there are terms in the progression; therefore 10 x 42=420 is the sum of both series, or twice the sum of the terms of the proposed progression. Whence the sum 10 × 42 required is 2 a, 210. Let us proceed in a similar manner with any A.P. :— INCREASING SERIES. a+d, a+2d, a+3d, a+4d, a+5d, a+6d, a+7d. DECREASING SERIES. a+7d, a+6d, a+5d, a+4d, a+3d, a+2d, a+d, a. 2a+7d, 2a+7d, 2a+7d, 2a+7d, 2a+7d, 2a+7d, 2a+7d, 2a +7d. In the first series every following term increases by d, whilst in the second it diminishes by d, The sum of the terms of both series is 8x (2a+7d), or the number of terms in one progression multiplied by the sum of any two corresponding terms; therefore 8x(2a+7d) the sum of the proposed series= 2 Were we to take the number of terms to be n, we should have: a, a+d, INCREASING SERIES. a+2d, a+3d...a+(n-2)d, a+(n-1)d. DECREASING SERIES. a+(n-1)d, a+(n-2)d, a+(n—3)d, a+(n—4)d,......a+d. a. 2a+(n-1)d, 2a+(n−1)d, &c........... .2a+(n−1)d. The sum of every two corresponding terms is 2a+(n−1)d, which quantity being repeated n times, we have, supposing s to (2a+ (n-1)d' represent the sum of the terms of one series, 2s=n· for the sum of both series; therefore the sum of the terms of any ·2a+ (nFrom which is derived the series is s=n. following rule: 2 |