172. If a certain number of terms of any arithmetical series be taken, the sum of the first and last terms is the same as that of any other two terms, 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 be, _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, as before. The rule, then, is: To sum any number of terms of an arithmetical progression, multiply the sum of the first and last terms by half the number of terms. For example, what are 99 terms of the series 1, 2, 3, &c.? The 99th term is 99, and the sum is 99 100 × 99 (99 + 1)---, or --------, or 4950. 2 2 The sum of 50 terms of the series 1 2 4 5 ( 1 50 ) 50 ---, ---, 1, ---, ---, 2, &c. is (--- + ---)---, 3 3 3 3 ( 3 3 ) 2 or 17 × 25, or 425. 173. The first term being given, and also the common difference and number of terms, the last term may be found by adding to the first term the common difference multiplied by one less than the number of terms. For it is evident that the second term differs from the first by the common difference, the _third_ term by _twice_, the _fourth_ term by _three_ times the common difference; and so on. Or, the passage from the first to the _n_th term is made by _n_-1 steps, at each of which the common difference is added. EXERCISES.