plzzzzzz Very Imp......among the natural numbers 1 to 49 find the numbers for which the sum of all preceding numbers and the sum of all succeeding numbers are equal
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Let the Sum be S
we know
Difference d = 1
from 1 to 30
S =n/2 [ a + (n-1)d ]
S = 30/2 [ 1 + (30-1)1 ]
S = 30 * 15
S = 450
from 31 to 49
S = 19/2 [ 1 + 18 ]
S = 450
Thus 1 to 30 and 31 to 49
we know
Difference d = 1
from 1 to 30
S =n/2 [ a + (n-1)d ]
S = 30/2 [ 1 + (30-1)1 ]
S = 30 * 15
S = 450
from 31 to 49
S = 19/2 [ 1 + 18 ]
S = 450
Thus 1 to 30 and 31 to 49
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