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Find the sum of the first 35 terms of the following series, to the nearest integer.

11,17,23,dots
Answer:

Find the sum of the first 3535 terms of the following series, to the nearest integer.\newline11,17,23, 11,17,23, \ldots \newlineAnswer:

Full solution

Q. Find the sum of the first 3535 terms of the following series, to the nearest integer.\newline11,17,23, 11,17,23, \ldots \newlineAnswer:
  1. Identify Terms: Identify the first term a1a_{1} and the common difference dd of the arithmetic series.\newlineThe first term a1a_{1} is 1111.\newlineTo find the common difference, subtract the first term from the second term: d=1711=6d = 17 - 11 = 6.
  2. Calculate Common Difference: Use the formula for the sum of the first nn terms of an arithmetic series: Sn=n2×(2a1+(n1)d)S_n = \frac{n}{2} \times (2a_1 + (n - 1)d). Here, n=35n = 35, a1=11a_1 = 11, and d=6d = 6.
  3. Use Sum Formula: Substitute the values into the formula to calculate the sum: S35=352×(2×11+(351)×6)S_{35} = \frac{35}{2} \times (2 \times 11 + (35 - 1) \times 6).
  4. Substitute Values: Perform the calculations inside the parentheses first: 2×11=222 \times 11 = 22 and (351)×6=34×6=204(35 - 1) \times 6 = 34 \times 6 = 204.
  5. Perform Calculations: Now, add the results inside the parentheses: 22+204=22622 + 204 = 226.
  6. Multiply Result: Multiply the result by n/2n/2: S35=352×226S_{35} = \frac{35}{2} \times 226.
  7. Calculate Final Sum: Calculate the sum: S35=17.5×226=3955S_{35} = 17.5 \times 226 = 3955.

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