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Find the sum of the first 7 terms of the following sequence. Round to the nearest hundredth if necessary.

125,quad100,quad80,dots
Sum of a finite geometric series:

S_(n)=(a_(1)-a_(1)r^(n))/(1-r)
Answer:

Find the sum of the first 77 terms of the following sequence. Round to the nearest hundredth if necessary.\newline125,100,80, 125, \quad 100, \quad 80, \ldots \newlineSum of a finite geometric series:\newlineSn=a1a1rn1r S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \newlineAnswer:

Full solution

Q. Find the sum of the first 77 terms of the following sequence. Round to the nearest hundredth if necessary.\newline125,100,80, 125, \quad 100, \quad 80, \ldots \newlineSum of a finite geometric series:\newlineSn=a1a1rn1r S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \newlineAnswer:
  1. Identify Sequence Type: To find the sum of the first 77 terms of the sequence, we first need to identify the type of sequence we are dealing with. The given sequence is 125125, 100100, 8080, ... which appears to be a geometric sequence because each term is obtained by multiplying the previous term by a constant ratio.
  2. Confirm Geometric Sequence: To confirm that this is a geometric sequence, we need to find the common ratio rr. We can do this by dividing the second term by the first term, and the third term by the second term. If the ratio is the same, then it is a geometric sequence.r=100125=0.8r = \frac{100}{125} = 0.8r=80100=0.8r = \frac{80}{100} = 0.8Since the ratio is the same, we have confirmed that this is a geometric sequence with a common ratio of 0.80.8.
  3. Apply Geometric Series Formula: Now that we have identified the sequence as geometric with a common ratio rr of 0.80.8, we can use the formula for the sum of the first nn terms of a geometric series:\newlineSn=(a1a1rn)(1r)S_n = \frac{(a_1 - a_1 \cdot r^n)}{(1 - r)}\newlinewhere SnS_n is the sum of the first nn terms, a1a_1 is the first term, and rr is the common ratio.
  4. Calculate Sum: We will now plug in the values into the formula to find the sum of the first 77 terms:\newlinea1=125a_1 = 125 (the first term)\newliner=0.8r = 0.8 (the common ratio)\newlinen=7n = 7 (the number of terms we want to sum)\newlineS7=125125×0.8710.8S_7 = \frac{125 - 125 \times 0.8^7}{1 - 0.8}
  5. Round to Nearest Hundredth: Let's calculate the sum:\newlineS7=125125×0.8710.8S_7 = \frac{125 - 125 \times 0.8^7}{1 - 0.8}\newlineS7=125125×0.20971520.2S_7 = \frac{125 - 125 \times 0.2097152}{0.2}\newlineS7=12526.21440.2S_7 = \frac{125 - 26.2144}{0.2}\newlineS7=98.78560.2S_7 = \frac{98.7856}{0.2}\newlineS7=493.928S_7 = 493.928
  6. Round to Nearest Hundredth: Let's calculate the sum:\newlineS7=125125×0.8710.8S_7 = \frac{125 - 125 \times 0.8^7}{1 - 0.8}\newlineS7=125125×0.20971520.2S_7 = \frac{125 - 125 \times 0.2097152}{0.2}\newlineS7=12526.21440.2S_7 = \frac{125 - 26.2144}{0.2}\newlineS7=98.78560.2S_7 = \frac{98.7856}{0.2}\newlineS7=493.928S_7 = 493.928Finally, we round the sum to the nearest hundredth as instructed:\newlineS7493.93S_7 \approx 493.93

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