What is the sum of seven odd consecutive natural numbers if the sum of the first, second

What is the sum of seven odd consecutive natural numbers if the sum of the first, second, fifth and seventh of them is 490.

Let us denote by x the first number from the given sequence of seven odd consecutive natural numbers.

Then the remaining six numbers from this sequence will be equal, respectively, x + 2, x + 4, x + 6, x + 8, x + 10 and x + 12, and the sum of all seven numbers will be:

x + x + 2 + x + 4 + x + 6 + x + 8 + x + 10 + x + 12 = 7x + 42.

According to the condition of the problem, the sum of the first, second, fifth and seventh numbers is 490, therefore, we can compose the following equation:

x + x + 2 + x + 8 + x + 12 = 490.

We solve the resulting equation:

4x + 22 = 490;

4x = 490 – 22;

4x = 468;

x = 468/4;

x = 117.

We find the sum of all seven numbers:

7x + 42 = 7 * 117 + 42 = 861.

Answer: the required amount is 861.