The sum of the terms of any A.P. is found by multiplying half the sum of the first and last terms by the number of terms in the series. 361. In this progression: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, we may observe that the sum of the first and last term is 42; the sum of the second and last but one is likewise 42; of the third and eighth; of the fourth and seventh, and so on of any two terms equally distant from the first and last. Also for any arithmetical series ; a, a+d, a+2d, a+3d, a+4d......a+(n—4)d, a+(n—3)d, a+(n-2)d, a +(n−1)d. The sum of the first and last term is equal to the sum of the second and last but one, &c., viz., 2a+(n−1)d. The reason of this is manifest from what has been said, for, we find the sum of two terms, as just mentioned, to be the same as adding together the two corresponding terms of an increasing and decreasing series. Therefore, if the sum of the first and last terms of a series, a+l, be multiplied by, half the number of terms, we obtain 2' = the sum of the terms of the series ; hence s=(a+1)2 = {2a+ (n-1)d} n 2 Ex. 1. In an arithmetical series, the first term a=4, the common difference d=5, and the number of terms n=36, find the sum s. Ex. 2. Given the first term a=3, the last term 1=36, and the number of terms n=18, to find the sum of the A.P.: 362. In the following propositions will be found the different cases which present themselves in A.P., they are deductions from what has been explained: 1. Given a, d, n, to find l. Ans. l=a+(n−1 d. 363. EXERCISES. 1. A debt was discharged by forty weekly payments in A.P., the first payment was £3 and the last £10. What is the common difference and the amount of the debt? 2. A gentleman bought a horse upon this condition, that for the first nail in its shoes he should pay 6d., for the second 10d., for the third 14d., and so on for every succeeding nail. Now the number of nails being 32, find the price of the horse. 3. A person travelled from Worksop to London, the distance being 148 miles. The first day he went 12 miles, and the last he went 25 miles, increasing his speed regularly every day. How long was he on his journey? 4. Suppose 100 stones were placed in a right line, a yard asunder and the first a yard from a basket, what distance will a man have travelled after he has brought them one by one to the basket? 5. Two persons, A and B, travel to meet each other. A goes at the uniform rate of 10 miles per hour. B goes 5 miles the first hour, and increases his speed 1 mile every hour. When they meet it is found that both have travelled the same distance. How many hours were they travelling, and how far were the two starting places from each other? 6. A workman served his master for 18 years. For the first year he received £20, and every successive year his salary was raised £3. What was the amount of his wages, and what did he receive for the last year? 7. What is the sum of all the numbers from 1 to 1000, both included? 8. A body falling in vacuo descends in the first second of time through a space of 16 feet, and in every successive second through 321 feet more. Through what space would it fall in 40 seconds, and how much would it descend the last second? 9. It is required to find eight arithmetical means between 1 and 1? 10. Twelve points being taken on the circumference of a circle, it is required to determine the greatest number of straight lines by which these points can be connected. 11. A carter takes 56 loads of stones, which are unloaded in as many heaps, equally distant from each other, on a road 1120 yards in length, the first heap is 250 yards from the quarry. What is the distance gone over by the carter in the performance of this undertaking? GEOMETRICAL PROGRESSION, OR 364. A series of numbers, the terms of which are continually increased or diminished by a multiplication or a division, is said to be in geometrical progression. Thus : 1 5 25 125 625 3125 15625 78125, &c. 4096 1024 256 64 16 4 1 16 54, &c. The first series being an increasing and the second a decreasing G.P. The terms of the former series are found by multiplying successively by 5, and those of the latter by dividing successively by 4. This constant multiplier or divisor is termed the ratio. If it be noticed that every term is deduced from the preceding › one by multiplying it by the ratio, these progressions might be put under the form of: 4' 42' 43' 1, 1×5, 1×52, 1×53, 1×54, 1× 55, 1×5°, 1 × 57, &c. 1 1 1 1x4o, 1×45, 1×44, 1×43, 1×42, 1×41, 1, &c. Or if the first term be expressed by a, the ratio by r, the number of terms by n, the geometrical series is represented thus: Here is plainly exhibited how any term of a G.P. is found by means of the first term and the ratio, viz., every term is the product of the first term by the ratio raised to the power of the number of terms less one. Thus the sixth term is the product of the first and of the ratio raised to the fifth power; the twelfth term is the product of the first by the ratio raised to the eleventh power; the last term, when the number of terms of the series is n, is the product of the first term by the ratio raised to the (n−1) power. Suppose to express the last term, then: 1=arn—1. 365. An important case in G P. is to determine the sum of the terms of the series. For this purpose let the terms of the G.P. be: 1, 5, 25, 125, 625, 3125, 15625, &c. Multiplying by the ratio 5, we obtain : 5, 25, 125, 625, 3125, 15625, 78125, &c. Subtracting the first series from the second, there remains 78125-1, which is equal to four times the sum of the terms of the proposed series, or to 4s (s representing the sum of the terms of the given G.P.) The same reasoning will apply to any other series, thus: a, ar, ar2, ar3, ar1, ar3......... ar-2, arm-1; multiplying by r, ar, ar2, ar3, art, ar3......arn—2, arn—1, arn. And the first series being taken from the second, leaves: sr—r=s(r—1)=arn_ -a. Now, dividing each side of this equality by (r-1), there remains : s= arn -a a(r*—1) r -1 Hence the sum of the terms of any geometrical series may be found by multiplying the first term by the ratio raised to the power of the number of terms, and dividing the difference between this product, and the first term by the ratio, less 1. 366. The several cases which may offer themselves in G.P. are presented in the following table; they are deductions from what has been investigated : 1. Given a, r, n, to find l. 2. Given a, r, s, to find l. 3. Given a, n, s, to find l. 4. Given r, n, s, to find l. 5. Given a, r, n, to find s. 6. Given a, r, l, to find s. Ans. s= |