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Find the sum of the first 6 terms of the sequence 3 , 6,12,dots A. 189 C. 141 B. 188 D. 187

99. Find the sum of the first 66 terms of the sequence 33 , 6,12,… 6,12, \ldots \newlineA. 189189\newlineC. 141141\newlineB. 188188\newlineD. 187187

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Q. 99. Find the sum of the first 66 terms of the sequence 33 , 6,12,… 6,12, \ldots \newlineA. 189189\newlineC. 141141\newlineB. 188188\newlineD. 187187
  1. Identify type of sequence: Identify the type of sequence. The sequence is geometric because each term is multiplied by a common ratio.
  2. Find common ratio: Find the common ratio. Common ratio = 63=2 \frac{6}{3} = 2
  3. Write sum formula: Write the formula for the sum of the first nn terms of a geometric sequence. Sum = a⋅rn−1r−1a \cdot \frac{r^n - 1}{r - 1} where aa = first term, rr = common ratio, nn = number of terms
  4. Plug in values: Plug in the values. a=3a = 3, r=2r = 2, n=6n = 6 Sum = 3imes26−12−13 imes \frac{2^6 - 1}{2 - 1}
  5. Calculate 262^6: Calculate 262^6. 26=642^6 = 64
  6. Calculate sum: Calculate the sum.\newlineSum = 3×(64−1)/13 \times (64 - 1) / 1\newlineSum = 3×633 \times 63\newlineSum = 189189

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