A Key Containing the Statements And Solutions of Questions in Prof Charles Davi
A Key Containing the Statements And Solutions of Questions in Prof Charles Davi
Charles Davies
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-u 1. 19 - 4 15 then we have, r = — - — =z -— — Si 5 5 hence, 4. 7. 10. 13. 16. 19, form the series. (8. ) First, to find the last term. We have, I = a — {71 — l)r. Make, a = 10, n = 21, and r — ^\ then, I =: 10 - (21 - l)\ == 10 - 6| = 3i; then, a /10 + 3J-\ ^^ 30, 10 ^^ 40 ^, 840, ^ 2 (9. ) We have the equations, S = y I X ^^, and I =z a + {7i — l)r. 270. ] AEITHMKTTCAL PKOGRKSSION. 69 In these equations all the quantities are known, except a and I, Substituting the numbers for the known quan...tities, we have, 2945 = (^-H!yx n, and 185 = a + {n - 1)6. ( 10. ) We have, from Art. 183, h — a m + 1 Making 5 = 5, and a =: 2, and m = 9, 5 2 we have, r = = 0. 3 ; from which the terms are easily found. (11. ) We have the formula, S = (-f-) X n. Now, (^ = 1, and I =: n; hence, (1 + n\ (n + 1\ ^=(-2-)>
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