find 3 digit number which are divisible 11. by AP method
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The sum of all 3 digits divisible by 11 is 44,550.
As we know that, the below numbers which are being divisible by 11, such that
110, 121, 132, 990
The above series looks like the number are in A.P
To find 'the last term of a series' is = a + (n-1) d.
Where a = 110, d = 11.
Such that,
990 = 110 + (n - 1)11
880 = (n – 1)11 80=n-1 n= 81
Sum of all the digits
= n/2 (2a + (n – 1)d)
81/2 (2(110)+80(11))
= 81/2(1100)
= 81 x 550 = 44550
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