find the sum of all three digit numbers which leave the remainder 3 when divided by 7 - Maths - Arithmetic Progressions

Question

find the sum of all three digit numbers which leave the
remainder 3 when divided by 7

Aakash Sharma · Accepted Answer

Dear Student!

Here is the answer to your query.

The three digits number which leave the remainder 3 when divided by 7 are

101, 108, 115, ..., 997

Clearly the sequence forms an A.P. with

First term (a) = 101

Common difference (d) = 108 – 101 = 7

and n^th term = 997

⇒ a + (n – 1)d = 997

⇒ 101 + (n – 1) 7 = 997

⇒ (n – 1) 7 = 997 – 101 = 896

$\Rightarrow (n - 1) = \frac{896}{7} = 128$

⇒ n = 128 + 1 = 129

Now

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{129} = \frac{129}{2} [2 \times 101 + (129 - 1) 7] = \frac{129}{2} [202 + 128 \times 7] = \frac{129}{2} [202 + 896] = \frac{129}{2} \times 1098 = 129 \times 549 = 70821$

Hence sum of all three digits number which leaves the remainder 3 when divided by 7 is 70821.

Cheers!

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