Let a_n be the n^th term of the series 5 + 8 + 14 + 23 + 35 + 50 +....and $S_{n} = \sum_{k = 1}^{n} a_{k}$ . Then S₃₀ - a₄₀ is equal to

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11310

11280

11260

11290

answer is C.

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Detailed Solution

$\begin{array}{l} S_{n} = 5 + 8 + 14 + 23 + 35 + 50 + \dots \dots + a_{n} \\ \frac{S_{n} = 5 + 8 + 14 + 23 + 35 + \dots . . . . . . . . . . . + a_{n}}{0 = 5 + 3 + 6 + 9 + 12 + 15 + . . . . . . . . . . . . . . - a_{n}} \end{array}$
$a_{n} = 5 + 3 + 6 + 9 + 12 + 15 + . . . . . . . . . . . . + a_{n - 1}$
$\begin{array}{ll} ∴ a_{n} = 5 + \frac{n - 1}{2} [6 + (n - 2) 3] \\ = 5 + \frac{n - 1}{2} [3 n] = 5 + \frac{3}{2} n^{2} - \frac{3}{2} n \\ = \frac{1}{2} [3 n^{2} - 3 n + 10] \\ ∴ a_{40} = 2345 \\ S_{n} = \frac{1}{2} [3 \sum n^{2} - 3 \sum n + 10 \sum 1] \\ S_{30} = \frac{1}{2} [\frac{3 \times 30 \times 31 \times 61}{6} - 3 \times \frac{30 \times 31}{2} + 10 \times 30] = 13, 635 \\ ∴ S_{30} - a_{40} = 11, 290 \end{array}$