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Find the Sum of All 3 - Digit Natural Numbers Which Are Divisible by 13. - Mathematics

Find the sum of all 3 - digit natural numbers which are divisible by 13.

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Solution

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.

= number of terms

All 3 digit natural number which is divisible by 13

So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.

Also, all these terms will form an A.P. with the common difference of 13.

So here,

First term (a) = 104

Last term (l) = 988

Common difference (d) = 13

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,

988 = 104 + (n - 1)13

988 = 104 + 13n - 13

988 = 91 + 13n

Further simplifying,

`n = (988 - 91)/13`

`n = 897/13`

n = 69

Now, using the formula for the sum of n terms, we get

`S_n = 69/2 [2(104) + (69 - 1)3]`

`= 69/2 [208 + (68)13]`

`= 69/2 (208 + 884)`

On further simplification, we get,

`S_n = 69/2 (1092)`

= 69 (546)

= 37674

Therefore, the sum of all the 3 digit multiples of 13 is `S_n = 37674`

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 12.3 | Page 51
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