Find the sum of the first 6 terms of the following sequence. Round to the nearest hundredth if necessary. 8,quad12,quad18,dots Sum of a finite geometric series: S_(n)=(a_(1)-a_(1)r^(n))/(1-r) Answer:

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Find the sum of the first 6 terms of the following sequence. Round to the nearest hundredth if necessary.  8,quad12,quad18,dots Sum of a finite geometric series:  S_(n)=(a_(1)-a_(1)r^(n))/(1-r) Answer:

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Identify sequence type: First, we need to identify the type of sequence we are dealing with. The given sequence is 8, 12, 18, ... which seems to be a geometric sequence where each term is multiplied by a common ratio to get the next term. To find the common ratio (r), we divide the second term by the first term. r = 12 / 8 = 1.5Calculate common ratio: Now that we have the common ratio, we can use the formula for the sum of the first n terms of a geometric series, which is S_n = (a_1 - a_1 * r^n) / (1 - r), where a_1 is the first term and n is the number of terms. We are looking for the sum of the first 6 terms, so n = 6 and a_1 = 8.Use sum formula: Let's plug the values into the formula and calculate the sum of the first 6 terms. S_6 = (8 - 8 * 1.5^6) / (1 - 1.5)Calculate 1.5^6: Now we calculate 1.5^6. 1.5^6 = 11.390625Substitute in formula: Substitute the value of 1.5^6 into the formula. S_6 = (8 - 8 * 11.390625) / (1 - 1.5)Calculate 8 * 11.390625: Now we calculate 8 * 11.390625. 8 * 11.390625 = 91.125Calculate numerator and denominator: Substitute the value into the formula. S_6 = (8 - 91.125) / (1 - 1.5)Divide numerator by denominator: Now we calculate the numerator and the denominator separately. Numerator: 8 - 91.125 = -83.125 Denominator: 1 - 1.5 = -0.5Perform final division: Finally, we divide the numerator by the denominator to find the sum of the first 6 terms. S_6 = -83.125 / -0.5Perform the division to get the final answer. S_6 = 166.25

Find the sum of the first $6$ terms of the following sequence. Round to the nearest hundredth if necessary. $\newline$ $8, \quad 12, \quad 18, \ldots$ $\newline$ Sum of a finite geometric series: $\newline$ $S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r}$ $\newline$ Answer:

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