Elements of Algebra

138 ELEMENTS SECT. III. wherefore the sum of the first and the last term is $3n-1$, and consequently the sum of this progression is equal to $\frac{n(3n-1)}{2} = \frac{3n^2-n}{2}$; and if we suppose $n=20$, the sum will be $10 \times 59 = 590$. 424. Again, let the first term be 1, the difference $d$, and the number of terms $n$; then the last term will be $1+(n-1)d$; to which adding the first, we have $2+(n-1)d$, and multiplying by the number of terms, we have $2n+n(n-1)d$; whence we deduce the sum of the progression $n + \frac{n(n-1)d}{2}$. And by making $d$ successively equal to 1, 2, 3, 4, &c., we obtain the following particular values, as shewn in the subjoined Table. If $d=1$, the sum is $n + \frac{n(n-1)}{2} = \frac{n^2+n}{2}$ $d=2$, - - - $n + \frac{2n(n-1)}{2} = n^2$ $d=3$, - - - $n + \frac{3n(n-1)}{2} = \frac{3n^2-n}{2}$ $d=4$, - - - $n + \frac{4n(n-1)}{2} = 2n^2-n$ $d=5$, - - - $n + \frac{5n(n-1)}{2} = \frac{5n^2-3n}{2}$ $d=6$, - - - $n + \frac{6n(n-1)}{2} = 3n^2-2n$ $d=7$, - - - $n + \frac{7n(n-1)}{2} = \frac{7n^2-5n}{2}$ $d=8$, - - - $n + \frac{8n(n-1)}{2} = 4n^2-3n$ $d=9$, - - - $n + \frac{9n(n-1)}{2} = \frac{9n^2-7n}{2}$ $d=10$, - - - $n + \frac{10n(n-1)}{2} = 5n^2-4n$ QUESTIONS FOR PRACTICE. * 1. Required the sum of an increasing arithmetical progression, having 3 for its first term, 2 for the common difference, and the number of terms 20. Ans. 440. 2. Required the sum of a decreasing arithmetical pro-