Question

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

Answer 1

104, 117,........988

a= 104
d= 13

So

988 = 104 + (n-1)13

884/13 +1 =n

69= n

So Sum = 69/2 ( 2× 104 + 68 × 13)

= 69/2 ( 208 + 884)

= 69/2 (1092)
= 37674

Answer 2

Answer:

37674

Step-by-step explanation:

(i)

The first three digit number divisible by 13 is 104.

After 104, to find the next 3 digit number divisible by 13, we have to add 13 to 104(104 + 13 = 117).So, the second 3 digit number divisible by 13 is 117.

In this way, to get the succeeding 3 digit numbers divisible by 13, we just have to add 13 as given below:

104,117,130,143...988.

Clearly, the above sequence of 3 digit numbers divisible by 13 forms an AP.

Here,

First term a = 104.

Common difference d = 117 - 104

                                     = 13.

Last term = tn = l = 988.

(ii)

We know that number of terms of an AP = a + (n - ) * d

⇒ 988 = 104 + (n - 1) * 13

⇒ 988 - 104 = (n -1 ) * 13

⇒ 884 = (n - 1) * 13

⇒ 884 = 13n - 13

⇒ 897 = 13n

⇒ n = 69.

(iii)

Sum of n terms of an AP = n/2[a + l]

⇒ 69/2[104 + 988]

⇒ 69/2[1092]

⇒ 69 * 546

⇒ 37674.

∴ The sum of all three-digit natural numbers divisible by 13 is 37674.

Hope this helps!