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Find the sum of all natural numbers lying between 100 and 200 which leave a remainder of 2 when divided by 5 in each case.

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Final answer:

The sum of all natural numbers lying between 100 and 200 which leave a remainder of 2 when divided by 5 is 2990.

Step-by-step explanation:

To find the sum of all natural numbers lying between 100 and 200 which leave a remainder of 2 when divided by 5, we need to identify all such numbers. These numbers form an arithmetic sequence starting from 102 (the smallest number greater than 100 that leaves a remainder of 2 when divided by 5) to 197 (the largest number less than 200 that leaves a remainder of 2 when divided by 5). The common difference of this sequence is 5 (since we are looking at every number that is 2 more than a multiple of 5).

The formula for the sum of an arithmetic sequence is given by S = n/2 × (a_1 + a_n), where n is the number of terms, a_1 is the first term, and a_n is the last term. To find n, we can use the formula n = ((a_n - a_1) / d) + 1, where d is the common difference.

In this case, a_1 = 102, a_n = 197, and d = 5. So, n = ((197 - 102) / 5) + 1 = 20. Now, we can calculate the sum: S = 20/2 × (102 + 197) = 10 × 299 = 2990.

The sum of all natural numbers lying between 100 and 200 which leave a remainder of 2 when divided by 5 is 2990.

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