4. If 100 stones be placed in a right line, exactly a yard asunder, and the first a yard from a basket, what length of ground will that man go, who gathers them up singly, returning with them one by one to the basket? Ans. 5 miles and 1300 yards. PROBLEM II. The first term, the last term, and the number of terms being given, to find the common difference. RULE.* Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common difference sought. EXAMPLES. 1. The extremes are 2 and 53, and the number of terms is 18; required the common difference. * The difference of the first and last terms evidently shews the increase of the first term by all the subsequent additions, till it becomes equal to the last; and' as the number of those additions is evidently one less than the number of terms, and the increase by every addition equal, it is plain, that the total increase, divided. by the number of additions, must give the difference at every one separately; whence the rule is manifest. 2. If the extremes be 3 and 19, and the number of terms 9, it is required to find the common difference, and the sum of the whole series. Ans. The difference is 2, and the sum is 99. 3. A man is to travel from London to a certain place in 12 days, and to go but 3 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 58 miles; required the daily increase, and the distance of the place from London. Ans. Daily increase 5, distance 366 miles. PROBLEM III. Given the first term, the last term, and the common difference, to find the number of terms. RULE. Divide the difference of the extremes by the common difference, and the quotient, increased by 1, is the num ber of terms required. EXAMPLES. *By the last problem, the difference of the extremes, divided by the number of terms less 1, gives the common difference; consequently the same, divided by the common difference, must give the number of terms less 1; hence this quotient, augmented by 1, must be the answer to the question. In any arithmetical progression, the sum of any two of its terms is equal to the sum of any other two terms, taken at an equal distance on contrary sides of the former; or the double of any one term is equal to the sum of any two terms, taken at an equal distance from it on each side. The sum of any number of terms (n) of the arithmetical series of odd numbers 1, 3, 5, 7, 9, &c. is equal to the square (n2) of that number. That EXAMPLES. 1. The extremes are 2 and 53, and the common difference 3; what is the number of terms? That is, if 1, 3, 5, 7, 9, &c. be the numbers, Then will 1, 22, 3, 4, 52, &c. be the sums of 1, 2, 3, &c, of those terms. For, o1 or the sum of I term = 1 or 1 9+7 or the sum of 4 terms = 42 or 16, &c. Whence it is plain, that, let n be any number whatsoever, the sum of n terms will be n2. The following table contains a summary of the whole doctrine of arithmetical progression. CASES OF ARITHMETICAL PROGRESSION. |