Geometric Sequence
Sum of Even- and Odd-Positioned Terms
🎯 What Does This Mean?
Sometimes we need the sum of terms only at even positions or only at odd positions in the sequence.
💡 The Key Idea
Key observation: Terms at even (or odd) positions form their own geometric sequence!
Terms at odd positions:
a₁, a₃, a₅, a₇, ...
First term: a₁
Common ratio: q²
Terms at even positions:
a₂, a₄, a₆, a₈, ...
First term: a₂ = a₁·q
Common ratio: q²
🔍 Why is the ratio q²?
Because jumping from one odd-position term to the next (or one even to the next) skips 2 positions.
Example: \(\frac{a_3}{a_1} = \frac{a_1 \cdot q^2}{a_1} = q^2\)
⚠️ Important: Number of Terms in Each Group
Depends on whether n (total number of terms) is even or odd!
n even (e.g. n = 8)
| Odd positions | \(\frac{n}{2}\) terms | 4 terms |
| Even positions | \(\frac{n}{2}\) terms | 4 terms |
n odd (e.g. n = 7)
| Odd positions | \(\frac{n+1}{2}\) terms | 4 terms |
| Even positions | \(\frac{n-1}{2}\) terms | 3 terms |
📐 The Formulas
Sum of terms at odd positions:
\(S_{\text{odd}} = \frac{a_1((q^2)^m - 1)}{q^2 - 1}\)
where m = number of terms at odd positions
Sum of terms at even positions:
\(S_{\text{even}} = \frac{a_1 \cdot q \cdot ((q^2)^m - 1)}{q^2 - 1}\)
where m = number of terms at even positions
✏️ Example 1: n Even
Question: Geometric sequence a₁ = 2, q = 3, n = 6.
Compute the sum of the even-positioned terms and the sum of the odd-positioned terms.
The sequence: a₁ = 2, a₂ = 6, a₃ = 18, a₄ = 54, a₅ = 162, a₆ = 486
Sum of odd-positioned terms (a₁, a₃, a₅):
Number of terms: m = 6/2 = 3
First term: a₁ = 2, ratio: q² = 9
\(S_{\text{odd}} = \frac{2(9^3 - 1)}{9 - 1} = \frac{2(729-1)}{8} = \frac{2 \cdot 728}{8} = \frac{1456}{8} = 182\)
Check: 2 + 18 + 162 = 182 ✓
Sum of even-positioned terms (a₂, a₄, a₆):
Number of terms: m = 6/2 = 3
First term: a₂ = 6, ratio: q² = 9
\(S_{\text{even}} = \frac{6(9^3 - 1)}{9 - 1} = \frac{6 \cdot 728}{8} = \frac{4368}{8} = 546\)
Check: 6 + 54 + 486 = 546 ✓
💡 Additional check:
Sum of all terms S₆ = 182 + 546 = 728
Compute directly: \(S_6 = \frac{2(3^6-1)}{3-1} = \frac{2 \cdot 728}{2} = 728\) ✓
✏️ Example 2: n Odd
Question: Geometric sequence a₁ = 1, q = 2, n = 7.
Compute the sum of the even-positioned terms.
The sequence: 1, 2, 4, 8, 16, 32, 64
Even positions: a₂ = 2, a₄ = 8, a₆ = 32
Solution:
Number of terms at even positions: m = (7−1)/2 = 3
First term: a₂ = 2, ratio: q² = 4
\(S_{\text{even}} = \frac{2(4^3 - 1)}{4 - 1} = \frac{2 \cdot 63}{3} = \frac{126}{3} = 42\)
Check: 2 + 8 + 32 = 42 ✓
🔗 Useful Relationship
\(\frac{S_{\text{even}}}{S_{\text{odd}}} = q\)
💡 Explanation: Every even-position term equals the preceding odd-position term multiplied by q.
Therefore the ratio of the sums is q (when each group has the same number of terms).
✏️ Verification from Example 1:
\(\frac{S_{\text{even}}}{S_{\text{odd}}} = \frac{546}{182} = 3 = q\) ✓
📝 Summary
Even/odd-positioned terms form a geometric sequence with ratio q²
Note the number of terms in each group!
Relationship: S_even / S_odd = q