provided one is as distant from the beginning of the series as the other is from the end. For example, let there be 7 terms, and let them
a b c d e f g.
Then, since, by the nature of the series, b is as much above a as f is below g (170), a + g = b + f . Again, since c is as much above b as e is below f (170), b + f = c + e . But a + g = b + f ; therefore a + g = c + e , and so on. Again, twice the middle term, or the term equally distant from the beginning and the end (which exists only when the number of terms is odd), is equal to the sum of the first and last terms; for since c is as much below d as e is above it, we have c + e = d + d = 2 d . But c + e = a + g ; therefore, a + g = 2 d . This will give a short rule for finding the sum of any number of terms of an arithmetical series. Let there be 7, viz. those just given. Since a + g , b + f , and c + e , are the same, their sum is three times ( a + g ), which with d , the middle term, or half a + g , is three times and a half ( a + g ), or the sum of the first and last terms multiplied by (3½), or ⁷/₂, or half the number of terms. If there had been an even number of terms, for example, six, viz. a , b , c , d , e , and f , we know now that a + f , b + e , and c + d , are the same, whence the sum is three times ( a + f ), or the sum of the first and last terms multiplied by half the number of terms,