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If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

Question

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

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Solution

To solve this problem, we need to use the formula for the sum of an arithmetic progression, which is S_n = n/2 * (a + l), where S_n is the sum of the first n terms, a is the first term, and l is the last term.

Given that the sum of the first 11 terms equals the sum of the first 19 terms, we can write the equation as follows:

11/2 * (a + l1) = 19/2 * (a + l2)

Solving for l1 and l2, we find that l1 = 19a - 11l2 and l2 = 11a - 19l1.

Substituting these values back into the equation, we get:

11/2 * (a + 19a - 11l2) = 19/2 * (a + 11a - 19l1)

Solving for a, we find that a = 0.

Since the first term is 0, this is a null sequence, and the sum of the first 30 terms is also 0.

This problem has been solved

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