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What is the sum of the first n terms of the arithmetic series 2, 5, 8, ...?
What is the sum of the first n terms of the arithmetic series 2, 5, 8, ...?
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Practice Questions
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Q1
What is the sum of the first n terms of the arithmetic series 2, 5, 8, ...?
n/2 * (2 + (n-1) * 3)
n * (2 + 3n)/2
3n^2/2 + n/2
n * (n + 1)
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The first term a = 2, common difference d = 3. The sum of the first n terms S_n = n/2 * (2a + (n-1)d) = n/2 * (2 + (n-1) * 3).
Questions & Step-by-step Solutions
1 item
Q
Q: What is the sum of the first n terms of the arithmetic series 2, 5, 8, ...?
Solution:
The first term a = 2, common difference d = 3. The sum of the first n terms S_n = n/2 * (2a + (n-1)d) = n/2 * (2 + (n-1) * 3).
Steps: 6
Show Steps
Step 1: Identify the first term of the series. The first term (a) is 2.
Step 2: Identify the common difference of the series. The common difference (d) is 5 - 2 = 3.
Step 3: Write the formula for the sum of the first n terms of an arithmetic series: S_n = n/2 * (2a + (n-1)d).
Step 4: Substitute the values of a and d into the formula: S_n = n/2 * (2*2 + (n-1)*3).
Step 5: Simplify the expression: S_n = n/2 * (4 + (n-1)*3).
Step 6: Further simplify: S_n = n/2 * (4 + 3n - 3) = n/2 * (3n + 1).
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