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The sum of all natural numbers lying between 100 and 1000 which are multiples of 5 is ____.

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Detailed Solution

The sum of all natural numbers lying between 100 and 1000 which are multiples of 5 is 98450.
We need to find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.
The numbers between 100 and 1000 which are multiples of 5 are 105, 110, 115…….,995 which is A.P. because the common difference is same.
We know,
The sum of the

n^{th}

term of the A.P is

S u m = n 2 2 a + n − 1 d

The formula for the

n^{th}

term of the A.P is

T n = a + n − 1 d

The first term is a.
The common difference is d.
The number of terms are n.
Therefore,

T n = a + n − 1 d ⇒ 995 = 105 + n − 1 5 ⇒ 995 − 105 = 5 n − 5890 ⇒ 5 n − 5

Now, according to the question,

890 = 5 n − 5 ⇒ 5 n = 890 + 5 ⇒ 5 n = 895 ⇒ n = 179

Therefore, the sum will be,

Sum = n 2 2 a + n − 1 d ⇒ Sum = 1792 2 × 105 + 179 − 1 5 ⇒ Sum = 1792 × 1100 ⇒ Sum = 98450

The sum of the series of a natural number between 100 and 1000 that are multiple of 5 is 98450

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