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Find the sum of the first 40 terms of the arithmetic sequence whose first term is −15 and whose common difference is 3.

User AHS
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Final answer:

To find the sum of the first 40 terms of an arithmetic sequence, you can use the formula: Sum = (n/2)(2a + (n-1)d), where n is the number of terms, a is the first term, and d is the common difference. Plugging in the values for n, a, and d given in the question, the sum is equal to 540.

Step-by-step explanation:

To find the sum of the first 40 terms of an arithmetic sequence, we can use the formula: Sum = (n/2)(2a + (n-1)d), where n is the number of terms, a is the first term, and d is the common difference. In this case, n = 40, a = -15, and d = 3. Plugging these values into the formula, we get:

Sum = (40/2)(2(-15) + (40-1)(3)) = (20)(-30 + 39)(3) = (20)(9)(3) = 540.

To determine the sum of the first 40 terms of an arithmetic sequence, the formula Sn = n/2[2a + (n-1)d] is utilized. In this formula, (n) denotes the number of terms, (a) signifies the first term, and (d) represents the common difference. Given that (n = 40), (a = -15), and (d = 3), the sum (S) is calculated as follows:

S = (40/2)[2(-15) + (40-1)(3)] = 20(-30 + 39)(3) = 540.

Hence, the sum of the first 40 terms in the arithmetic sequence is 540.

User SomeGuyOnAComputer
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