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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

Find the sum of even numbers between 50 and 100

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

1 Answer

Toolbox:
  • 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 $

 

answered Nov 19, 2013 by harini.tutor
 

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