Question

FIND THE SUM OF ALL THREE DIGIT POSITIVE INTEGERS WHICH ARE DIVISIBLE BY 7

Answer 1

first find the number of terms and then substitute it In the formula of Sn

Answer 2

Answer:

The sum of all three digit positive integers which are divisibility 7 is 70336.

Step-by-step explanation:

The smallest and the largest three-digit number, which are divisible by 7 are 105 and 994 respectively.

So, the sequence of three-digit numbers which are divisible by 7 is 105,112,119,…,994

Clearly, it is an A.P. with first term a = 105

and common difference d =7.Let there be n terms in this sequence.

Then,

$a_n = 994$

Or,

$a + (n - 1) \frac{d}{2} = 994$

So,

$105 + (n - 1) \frac{7}{2} = 994 \\ (n - 1) \frac{7}{2} = 994 - 105 \\ (n - 1) \frac{7}{2} = 889 \\ (n - 1) = 889 \times \frac{2}{7} \\ (n - 1) = 127 \\ n = 127 + 1 \\ n = 128$

So,there are total 128 positive three digit numbers which are divisible by 7.

We know,

$s_n = \frac{n}{2} (a + l)$

Where

$s_n$ is sum of all terms and n is number of terms and a is first term of the series and l is last term of the series.

So,

$s_n = \frac{128}{2} (105 + 994) \\ = 64 \times 1099 \\ = 70336$

Required sum is 70336.

Here I used rule of Arithmetic progression.

Know more about Arithmetic progression:

1) https://brainly.in/question/4219484

2) https://brainly.in/question/2768711

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