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Find the sum of the first 38 terms of the following series, to the nearest integer. 9,12,15,dots Answer:

Find the sum of the first 3838 terms of the following series, to the nearest integer.\newline9,12,15,… 9,12,15, \ldots \newlineAnswer:

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Full solution

Q. Find the sum of the first 3838 terms of the following series, to the nearest integer.\newline9,12,15,… 9,12,15, \ldots \newlineAnswer:
  1. Identify type and difference: Identify the type of series and the common difference.\newlineThe given series is arithmetic because there is a constant difference between consecutive terms.\newlineTo find the common difference dd, subtract the first term from the second term.\newlined=12−9=3d = 12 - 9 = 3
  2. Find first term and number: Find the first term aa and the number of terms nn. The first term aa is given as 99. The number of terms nn to sum up is given as 3838.
  3. Use sum formula: Use the formula for the sum of the first nn terms of an arithmetic series.\newlineThe formula is Sn=n2×(2a+(n−1)d)S_n = \frac{n}{2} \times (2a + (n - 1)d).\newlineNow plug in the values of aa, nn, and dd into the formula.\newlineS38=382×(2×9+(38−1)×3)S_{38} = \frac{38}{2} \times (2\times9 + (38 - 1)\times3)
  4. Simplify expression: Simplify the expression.\newlineS38=19×(18+37×3)S_{38} = 19 \times (18 + 37\times3)\newlineS38=19×(18+111)S_{38} = 19 \times (18 + 111)\newlineS38=19×129S_{38} = 19 \times 129
  5. Calculate sum: Calculate the sum.\newlineS38=19×129S_{38} = 19 \times 129\newlineS38=2451S_{38} = 2451

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