(a) 1800 (b) 2800 (c) 5800 (d) 5000

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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 = 52, \quad d = 2, \quad T_{n} = 98 $

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

$\therefore 98 = 52 + (n - 1)2 \implies 42 = (n - 1)2 \implies n = 24 $

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

$ S_{n} = \large \frac {24}{2} [2 X 52 + (24 - 1) X 2] $

$ S_{n} = 1800 $

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