Geometric Sequence
Finding the General Term from the Sum Formula
💡 The Key Idea
Sometimes we are given a formula for Sₙ of some sequence and asked to find the general term aₙ.
The relationship between sum and term:
Sₙ = sum of the first n terms = a₁ + a₂ + … + aₙ
Sₙ₋₁ = sum of the first (n−1) terms = a₁ + a₂ + … + aₙ₋₁
The difference:
Sₙ − Sₙ₋₁ = aₙ (the n-th term)
⭐ Formula for Finding aₙ from Sₙ
\(a_n = S_n - S_{n-1}\)
(for n ≥ 2)
⚠️ Very important!
The formula is valid only for n ≥ 2
a₁ a₁ is found by substituting n = 1 into the sum formula: \(a_1 = S_1\)
📋 Solution Steps
Find a₁: substitute n = 1 into Sₙ to get S₁ = a₁
Find Sₙ₋₁: in the Sₙ formula, replace every n with (n−1)
Compute aₙ: calculate aₙ = Sₙ − Sₙ₋₁ and simplify
Verify: confirm the formula gives a₁ when n = 1
✏️ Example 1 (Detailed)
Given: \(S_n = 3^n - 1\)
Find: the general term aₙ and check whether this is a geometric sequence.
Step 1 – Finding a₁:
\(a_1 = S_1 = 3^1 - 1 = 3 - 1 = 2\)
Step 2 – Finding Sₙ₋₁:
Replace n with (n−1):
\(S_{n-1} = 3^{n-1} - 1\)
Step 3 – Computing aₙ (for n ≥ 2):
\(a_n = S_n - S_{n-1}\)
\(a_n = (3^n - 1) - (3^{n-1} - 1)\)
\(a_n = 3^n - 1 - 3^{n-1} + 1\)
\(a_n = 3^n - 3^{n-1}\)
\(a_n = 3^{n-1}(3 - 1)\)
\(a_n = 2 \cdot 3^{n-1}\)
Step 4 – Verification:
Substitute n = 1 into the formula we found:
\(a_1 = 2 \cdot 3^{1-1} = 2 \cdot 3^0 = 2 \cdot 1 = 2\) ✓
Matches! So the formula \(a_n = 2 \cdot 3^{n-1}\) is valid for all n ≥ 1
Is this a geometric sequence?
The formula \(a_n = 2 \cdot 3^{n-1}\) is exactly the general term formula of a geometric sequence!
Yes! It is a geometric sequence with a₁ = 2 and q = 3
✏️ Example 2
Given: \(S_n = 5 \cdot 2^n - 5\)
Find: the general term aₙ.
Step 1: \(a_1 = S_1 = 5 \cdot 2^1 - 5 = 10 - 5 = 5\)
Step 2: \(S_{n-1} = 5 \cdot 2^{n-1} - 5\)
Step 3:
\(a_n = S_n - S_{n-1} = (5 \cdot 2^n - 5) - (5 \cdot 2^{n-1} - 5)\)
\(= 5 \cdot 2^n - 5 - 5 \cdot 2^{n-1} + 5\)
\(= 5 \cdot 2^n - 5 \cdot 2^{n-1}\)
\(= 5 \cdot 2^{n-1}(2 - 1)\)
\(= 5 \cdot 2^{n-1}\)
Step 4: Verify: \(a_1 = 5 \cdot 2^0 = 5\) ✓
Answer: \(a_n = 5 \cdot 2^{n-1}\) (geometric sequence with a₁ = 5, q = 2)
⚠️ Special Case – When the Formula Is Invalid for n = 1
Example: \(S_n = 2^n + 1\)
Step 1: \(a_1 = S_1 = 2^1 + 1 = 3\)
Step 2: \(S_{n-1} = 2^{n-1} + 1\)
Step 3:
\(a_n = (2^n + 1) - (2^{n-1} + 1) = 2^n - 2^{n-1} = 2^{n-1}(2-1) = 2^{n-1}\)
Step 4 – Verification:
Substitute n = 1: \(2^{1-1} = 2^0 = 1\)
But a₁ = 3, not 1! ❌
💡 In such cases write:
\(a_n = \begin{cases} 3 & n = 1 \\ 2^{n-1} & n \geq 2 \end{cases}\)
⚠️ Note: Such a sequence is not a geometric sequence because the first term "breaks" the pattern!
📝 Summary
\(a_n = S_n - S_{n-1}\) (for n ≥ 2)
\(a_1 = S_1\)
Always check whether the formula is also valid for n = 1!