Arithmetic & Geometric Series
Precalculus
Intro: Compute partial sums from concrete parameters.
Worked example
- Find $S_{10}$ for $a_1=3,\,d=2$
- Goal: Compute the partial sum $S_{10}$ of an arithmetic sequence with first term $a_1=3$ and common difference $d=2$.
- Formula (version 1): $\displaystyle S_n = \frac{n}{2}(a_1 + a_n)$.
- First find the 10th term: $a_{10} = a_1 + (10-1)d = 3 + 9\cdot 2 = 3 + 18 = 21$.
- Plug into the sum formula: $\displaystyle S_{10} = \frac{10}{2}(3 + 21) = 5\cdot 24 = 120$.
- Check with equivalent formula (version 2): $\displaystyle S_n = \frac{n}{2}\big(2a_1 + (n-1)d\big)$.
- Compute: $\displaystyle S_{10} = \frac{10}{2}\big(2\cdot 3 + 9\cdot 2\big) = 5\big(6 + 18\big) = 5\cdot 24 = 120$.
- Conclusion: $\boxed{S_{10} = 120}$.
- Find $S_{6}$ for $a_1=5,\,r=\tfrac{1}{2}$
- Goal: Compute the partial sum $S_6$ of a geometric sequence with first term $a_1=5$ and ratio $r=\tfrac{1}{2}$.
- Formula (for $r\neq 1$): $\displaystyle S_n = a_1\,\frac{1 - r^n}{1 - r}$.
- Compute $r^6$: $\left(\tfrac{1}{2}\right)^6 = \tfrac{1}{64}$.
- Substitute values: $\displaystyle S_6 = 5\,\frac{1 - \tfrac{1}{64}}{1 - \tfrac{1}{2}}$.
- Simplify numerator: $1 - \tfrac{1}{64} = \tfrac{63}{64}$.
- Simplify denominator: $1 - \tfrac{1}{2} = \tfrac{1}{2}$.
- Divide by $\tfrac{1}{2}$ (i.e., multiply by $2$): $\displaystyle S_6 = 5 \cdot \tfrac{63}{64} \cdot 2 = 10 \cdot \tfrac{63}{64} = \tfrac{630}{64}$.
- Reduce the fraction by $2$: $\tfrac{630}{64} = \tfrac{315}{32}$.
- Decimal form (optional): $\tfrac{315}{32} = 9.84375$.
- Conclusion: $\boxed{S_6 = \tfrac{315}{32}\;\,(=9.84375)}$.
FAQs
Convergence?
Infinite geometric series need $|r|<1$.
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