find the sum of all three digit number which are divisible by 11
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The first three digit no. which is divisible by 11 is 110
a (first term) = 110
d (common difference) = 11
an (last term) = 990
an = a + (n-1)d
990 = 110 + (n-1)11
990 - 110 = (n-1)11
880/11 = (n-1)
80 = n - 1
n = 81
So the no. of terms are 81...Putting all values in Sn
Sn = n/2 [2a + (n-1)d]
= 81/2 [2*110 + (81-1)(11)]
= 81/2 [220 + 80*11}
= 81/2 [220 + 880)
= 81/2*1100
= 81*550
= 44550
a (first term) = 110
d (common difference) = 11
an (last term) = 990
an = a + (n-1)d
990 = 110 + (n-1)11
990 - 110 = (n-1)11
880/11 = (n-1)
80 = n - 1
n = 81
So the no. of terms are 81...Putting all values in Sn
Sn = n/2 [2a + (n-1)d]
= 81/2 [2*110 + (81-1)(11)]
= 81/2 [220 + 80*11}
= 81/2 [220 + 880)
= 81/2*1100
= 81*550
= 44550
kanamsahani:
but 987 is also divisible by 11
nope , it isn't 987÷11
=89.7272727273
=89.7272727273
sorry 979
979+11
=990
=990
990 = an (last term)
ohk thanks
wlcm :)
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