The sum of all the three digit natural numbers which are multiples of 7 is ____.

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Detailed Solution

The sum of all the three digit natural numbers which are multiples of 7 is 70336.
We need to find the sum of all the three digit natural numbers which are multiples of 7.
We know,
The formula of the sum of first n terms of an A.P is,

S n = n 2 2 a 1 + n − 1 d

S n = n 2 a 1 + L

Where,
The first term, a=105.
The common difference is d.
The number of terms are n.
According to the question,

A . P = 105, 112, 119, 126, ............994

The first term of the series is

a_{1} = 105 .

The last term of the series is

a_{n} = 994 .

The common difference of the series is,

d = 112 − 105 ⇒ d = 7

Therefore,

a n = a 1 + n − 1 d ⇒ 994 = 105 + n − 1 7 ⇒ 994 − 105 = n − 1 7 ⇒ 8897 + 1 = n ⇒ n = 889 + 7 7 ⇒ n = 8967 ⇒ n = 128

So, there is a total of 128 terms in the A.P.
Now, the required sum will be,

S n = n 2 a 1 + L ∵ a n = L ⇒ S 128 = 1282 105 + 994 ⇒ S 128 = 70336

So, the sum of all three-digit natural numbers which are multiples of 7 is 70336.

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