Additional Mathematics – Series | e-Consult
Series (1 questions)
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Answer 3
- The convergence condition is \(|r|
- Using \(S_{\infty}= \dfrac{a}{1-r}\):
\(S_{\infty}= \dfrac{12}{1-(-\frac{2}{5})}= \dfrac{12}{1+\frac{2}{5}}= \dfrac{12}{\frac{7}{5}}= \frac{12 \times 5}{7}= \frac{60}{7}\approx 8.57.\)
- The sum of the first \(n\) terms is \(S_n = a\frac{1-r^{\,n}}{1-r}\).
We need \(S_n > 10\). Compute iteratively:
| \(n\) | \(S_n\) (exact) | \(S_n\) (decimal) |
| 1 | \(12\) | 12.00 |
| 2 | \(12\frac{1-(-\frac{2}{5})^{2}}{1+\frac{2}{5}} = 12\frac{1-\frac{4}{25}}{\frac{7}{5}} = 12\frac{\frac{21}{25}}{\frac{7}{5}} = 12\cdot\frac{21}{25}\cdot\frac{5}{7}=12\cdot\frac{3}{5}=7.2\) | 7.20 |
| 3 | \(12\frac{1-(-\frac{2}{5})^{3}}{1+\frac{2}{5}} = 12\frac{1+\frac{8}{125}}{\frac{7}{5}} = 12\frac{\frac{133}{125}}{\frac{7}{5}} = 12\cdot\frac{133}{125}\cdot\frac{5}{7}=12\cdot\frac{133}{175}= \frac{1596}{175}\approx 9.12\) | 9.12 |
| 4 | \(12\frac{1-(-\frac{2}{5})^{4}}{1+\frac{2}{5}} = 12\frac{1-\frac{16}{625}}{\frac{7}{5}} = 12\frac{\frac{609}{625}}{\frac{7}{5}} = 12\cdot\frac{609}{625}\cdot\frac{5}{7}=12\cdot\frac{609}{875}= \frac{7308}{875}\approx 8.36\) | 8.36 |
Since \(S_1 = 12 > 10\), the smallest integer \(n\) satisfying the condition is \(n = 1\).