find the sum of all 3 -digit numberswhich are divisible by 11
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2
Hi !
The no:s divisible by 11 forms an A.P :-
110,121,132, ........ ,990.
a = 110
d = 11
an = 990
n = ?
an = a + (n - 1) d
990 = 110 + (n - 1)11
880 = (n - 1) 11
(n - 1) = 80
n = 81
Sn = n/2 ( a + an)
= 81/2 * ( 110 + 990)
= 81/2*1100
= 44550
Sum of all three digit no:s divisible by 11 is 44550
The no:s divisible by 11 forms an A.P :-
110,121,132, ........ ,990.
a = 110
d = 11
an = 990
n = ?
an = a + (n - 1) d
990 = 110 + (n - 1)11
880 = (n - 1) 11
(n - 1) = 80
n = 81
Sn = n/2 ( a + an)
= 81/2 * ( 110 + 990)
= 81/2*1100
= 44550
Sum of all three digit no:s divisible by 11 is 44550
Answered by
0
The first three digit term divisible by 11 is 110
So first term a = 110
d (common difference) = 11
and last term tn=990
From the formula tn=a+(n-1)d, we get n=81
So the number of terms is 81. Now we put this value in the formula
Sum=n/2[2a+(n-1)d]
Then on solving we get sum=44550
HOPE THIS HELPS!
So first term a = 110
d (common difference) = 11
and last term tn=990
From the formula tn=a+(n-1)d, we get n=81
So the number of terms is 81. Now we put this value in the formula
Sum=n/2[2a+(n-1)d]
Then on solving we get sum=44550
HOPE THIS HELPS!
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