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306 ALGEBRA. Ex. 2. The sum of three numbers in A. P. is 33, and their product is 792; find them.

Let a be the middle number, d the common difference; then the three numbers are a - d, a, a + d. Hence a — d + a + a + d = 33; whence a = 11; and the three numbers are 11 — d, 11, 11 + d. 11(11 + d)(11 - d)= 792, 121 - d2 = 72, d = ± 7; and the numbers are 4, 11, 18.

Ex. 3. How many terms of the series 24, 20, 16,... must be taken that the sum may be 72 ?

Let the number of terms be n; then, since the common difference is 20 - 24, or - 4, we have from (3), Art. 367,

72 = n/2 {2x24+(n-l)(-4)} = -24n - 2n(n- 1) ; whence n2 - 13 n + 36 = 0, or (n-4)(n-9)=0; n = 4 or 9.

Both of these values satisfy the conditions of the question; for if we write down the first 9 terms, we get 24, 20, 16, 12. 8, 4, 0, - 4, - 8 ; and, as the last five terms destroy each other, the sum of 9 terms is the same as that of 4 terms.

Ex. 4. An A. P. consists of 21 terms; the sum of the three terms in the middle is 129, and of the last three is 237; find the series.

Let a be the first term, and d the common difference. Then 237 = the sum of the last three terms = a + 20 d + a + 19 d + a + 18 d = 3 a + 57 d; whence a + 19d = 79. (1) Again, the three middle terms are the 10th, 11th, 12th; hence 129 = the sum of the three middle terms = a + 9d + a + 1Od + a + 11d = 3a + 30d; whence a + 10d = 43 (2). From (1) and (2), we obtain d = 4, a = S. Hence the series is 3, 7, 11, ..., 83.