Question

find the sum of all 3 digit natural numbers divisible by 13 using arithmetic progession​

Answer 1

Step-by-step explanation:

The smallest and the largest numbers of three digits, which are divisible by 13 are 104 and 988 respectively.

So, the sequence of three digit numbers which are divisible by 13 are 104,117,130,...,988.

It is an AP with first term a=104 and common difference d=13.

Let there be n terms in this sequence. Then,

a

n

=988⇒a+(n−1)d=988⇒104+(n−1)×13=988⇒n=69

Now, Required sum =

2

n

[2a+(n−1)d]

=

2

69

[2×104+(69−1)×13]

=

2

69

[1092]

=69×546=37674

Answer 2

Answer:

AP - 104,117,.....................988

So, a = 104, d =13, last term(l) = 988

. . . . Last term = a + ( n - 1)d = 988

. . . . . . . . . . . 104 + (n-1)*13 =988

. . . . . . . . . . . (n-1)*13 = 988 - 104 = 884

. . . . . . . . . . . n-1 = 884 / 13

. . . . . . . . . . . n-1 = 68

. . . . . . . . . . . n = 68 +1 =69

. . . . . . . n = 69

. . . Sum of all 3 digits no. divisible by 13 = n/2 ( a + l)

. . . . . . . . . . . . . . = 69 /2 ( 104 + 988)

. . . . . . . . . . . . . . = 69 * 1092 /2 = 69 * 546

. . . . . . . . Sum of AP = 37674