(a) 18000 (b) 5000 (c) 10000 (d) 19900

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- Sum of an AP is given by $ S_{n} = \large \frac {n}{2} [2a + (n-1)d] $
- nth term of an AP is given by $ T_{n} = \large a + (n-1)d $

Sum of an AP is given by $ S_{n} = \large \frac {n}{2} [2a + (n-1)d] $

Here $ a = 1, \quad d = 2, \quad T_{n} = 199 $

$ T_{n} = a + (n-1)d $

$\therefore 199 = 1 + (n - 1)2 \implies 198 = (n - 1)2 \implies n = 100 $

$ S_{n} = \large \frac {n}{2} [2a + (n-1)d] $

$ S_{n} = \large \frac {100}{2} [2 X 1 + (100 - 1) X 2] $

$ S_{n} = 10000 $

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