find the sum of all integers between 92 and 76 , which are multiples of 9
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4
Answer:
In between 92 and 786, The first multiple of 9 is 99 and the last multiple is 783.
Therefore the AP is 99, 108, ...... 783
a = 99, l = 783, d = 9
l = a + (n - 1)d
783 = 99 + (n-1)9
By solving, you get, n = 77
Therefore, the sum = n/2 [ a + l ]
= 77/2 [ 99 + 783 ]
= 77/2 ( 882 )
= 33957
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