Question

Show that \(a_1, a_2, . . ., a_n, .....\) form an AP where an is defined as below:

1. \(a_n = 3 + 4n\)
2. \(a_n = 9 – 5n\)

Also find the sum of the first 15 terms in each case.

Answer 1

(i) Given the sequence defined by \(a_n = 3 + 4n\): \(a_1 = 3 + 4(1) = 7\) \(a_2 = 3 + 4(2) = 11\) \(a_3 = 3 + 4(3) = 15\) \(a_4 = 3 + 4(4) = 19\) The differences between consecutive terms are: \(a_2 - a_1 = 11 - 7 = 4\) \(a_3 - a_2 = 15 - 11 = 4\) \(a_4 - a_3 = 19 - 15 = 4\) Since the difference between consecutive terms (\(a_{k+1} - a_k\)) is constant, this is an arithmetic progression (AP) with a common difference \(d = 4\) and the first term \(a = 7\). The sum of the first \(n\) terms of an AP is given by \(S_n = \frac n2 [2a + (n-1)d]\). We calculate the sum of the first 15 terms: \(S_{15} = \frac {15}{2} [2(7) + (15-1)4]\) \(S_{15} = \frac {15}{2} [14 + 14(4)]\) \(S_{15} = \frac {15}{2} [14 + 56]\) \(S_{15} = \frac {15}{2} (70)\) \(S_{15} = 15 \times 35\) \(S_{15} = 525\)

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(ii) Given the sequence defined by \(a_n = 9 - 5n\): \(a_1 = 9 - 5 \times 1 = 4\) \(a_2 = 9 - 5 \times 2 = -1\) \(a_3 = 9 - 5 \times 3 = -6\) \(a_4 = 9 - 5 \times 4 = -11\) The differences between consecutive terms are: \(a_2 - a_1 = -1 - 4 = -5\) \(a_3 - a_2 = -6 - (-1) = -5\) \(a_4 - a_3 = -11 - (-6) = -5\) Since the difference between consecutive terms (\(a_{k+1} - a_k\)) is constant, this is an arithmetic progression (AP) with a common difference \(d = -5\) and the first term \(a = 4\). The sum of the first \(n\) terms of an AP is given by \(S_n = \frac n2 [2a + (n-1)d]\). We calculate the sum of the first 15 terms: \(S_{15} = \frac {15}{2} [2(4) + (15-1)(-5)]\) \(S_{15} = \frac {15}{2} [8 + 14(-5)]\) \(S_{15} = \frac {15}{2} [8 - 70]\) \(S_{15} = \frac {15}{2} (-62)\) \(S_{15} = 15 (-31)\) \(S_{15} = -465\)