Given a + 2d = 12 and a + 49d = 106, solving these simultaneous equations yields first term a = 8 and common difference d = 2. Therefore, the 29th term evaluates to 8 + 28 x 2 = ...
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In this sequence, the first term is 11 and common difference is -3. The explicit nth term simplifies to tn = 14 – 3n, while its recursive rule is stated as t1 = 11, tn = tn-1 – 3 for ...
Here, first term is 21 and common difference is -3. Solving tn = a + (n – 1)d shows that -81 is the 35th term, and 0 is indeed the 8th term since n is a positive integer.
For the given arithmetic progression, the first term is 3 and the common difference is 5. Using the formula tn = a + (n – 1)d, the 10th term is 48 and the 26th term is 128.
For any chord through A, its length depends on its distance from O. The chord is shortest when this distance is greatest. This occurs when the chord is perpendicular to OA. Hence proved.