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प्रश्न
Find the sum of all 3-digit natural numbers, which are multiples of 11.
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उत्तर
In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
`S_n = n/2 [2a + (n - 1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
all 3-digit natural numbers, which are multiples of 11.
We know that the first 3 digit number multiple of 11 will be 110.
Last 3 digit number multiple of 11 will be 990.
So here,
First term (a) = 110
Last term (l) = 990
Common difference (d) = 11
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
`a_n = a + (n - 1)d`
So for the last term
990 = 110 + (n -1) 11
990 = 110 + 11n - 11
990 = 99 + 11n
891 = 11n
81 = n
Now, using the formula for the sum of n terms, we get
`S_n = 81/2 [2(110) + (81 - 1)11]`
`S_n = 81/2 [220 + 80 xx 11]`
`S_n = 81/2 xx 1100`
`S_n = 81 xx 550`
`S_n = 44550`
Therefore, the sum of all the 3 digit multiples of 11 is 44550.
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