Find the sum of the first 11 positive numbers which are multiples of 6.
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Sn = n\2 [ 2a + ( n - 1 )d ]
= 11\2 [ 2*6 + ( 11 - 1 )6 ] (since 6 is the first positive multiple and on adding 6 the multiples go on )
= 11\2 [ 12 + 60 ]
= 11\2 * 72
= 11 * 36
= 396
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Answer:
The sum of first 11 positive numbers which are multiple of 6 is 396
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