Elements of Algebra

CHAP. IV. OF ALGEBRA. 135 CHAP. IV. *Of the Summation of Arithmetical Progressions.* 412. It is often necessary also to find the sum of an arithmetical progression. This might be done by adding all the terms together; but as the addition would be very tedious, when the progression consisted of a great number of terms, a rule has been devised, by which the sum may be more readily obtained. 413. We shall first consider a particular given progression, in which the first term is 2, the difference 3, the last term 29, and the number of terms 10; 1 2 3 4 5 6 7 8 9 10 2, 5, 8, 11, 14, 17, 20, 23, 26, 29. In this progression, we see that the sum of the first and last term is $31$; the sum of the second and the last but one $31$; the sum of the third and the last but two $31$, and so on: hence we conclude, that the sum of any two terms equally distant, the one from the first, and the other from the last, is always equal to the sum of the first and the last term. 414. The reason of this may be easily traced; for if we suppose the first to be $a$, the last $z$, and the difference $d$, the sum of the first and the last term is $a+z$; and the second term being $a+d$, and the last but one $z-d$, the sum of these two terms is also $a+z$. Farther, the third term being $a+2d$, and the last but two $z-2d$, it is evident that these two terms also, when added together, make $a+z$; and the demonstration may be easily extended to any other two terms equally distant from the first and last. 415. To determine, therefore, the sum of the progression proposed, let us write the same progression term by term, inverted, and add the corresponding terms together, as follows: $2+5+8+11+14+17+20+23+26+29$ $29+26+23+20+17+14+11+8+5+2$ -------------------------------------------------- $31+31+31+31+31+31+31+31+31+31$ This series of equal terms is evidently equal to twice the sum of the given progression: now, the number of those