If the sum of first n terms of a series is (n + 12) , then what is its third term?
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Understanding the Problem: Finding the Third Term
The question asks us to find the value of the third term in a series, given a formula for the sum of its first \(n\) terms. The sum of the first \(n\) terms is denoted by \(S_n\), and we are given that \(S_n = n + 12\).
To find any term of the series, say the \(n\)-th term (\(a_n\)), we can use the relationship between the sum of the first \(n\) terms and the sum of the first \(n-1\) terms. The sum of the first \(n\) terms (\(S_n\)) includes the \(n\)-th term, while the sum of the first \(n-1\) terms (\(S_{n-1}\)) does not. Therefore, the \(n\)-th term can be found by subtracting the sum of the first \(n-1\) terms from the sum of the first \(n\) terms:
\(a_n = S_n - S_{n-1} \quad \text{for } n > 1\)
For the first term (\(a_1\)), the sum of the first term is simply the term itself, so \(a_1 = S_1\).
Step-by-Step Calculation
We need to find the third term, \(a_3\). Using the formula \(a_n = S_n - S_{n-1}\), we can find \(a_3\) if we know \(S_3\) and \(S_2\). We can calculate \(S_3\) and \(S_2\) using the given formula \(S_n = n + 12\).
Step 1: Find the sum of the first 1 term (\(S_1\)) and the first term (\(a_1\)).
Using the formula \(S_n = n + 12\), substitute \(n=1\):
\(S_1 = 1 + 12 = 13\)
The sum of the first term is the first term itself. So, \(a_1 = S_1 = 13\).
Step 2: Find the sum of the first 2 terms (\(S_2\)).
Using the formula \(S_n = n + 12\), substitute \(n=2\):
\(S_2 = 2 + 12 = 14\)
Step 3: Find the second term (\(a_2\)).
Using the relationship \(a_n = S_n - S_{n-1}\), for \(n=2\), we have \(a_2 = S_2 - S_1\). We already calculated \(S_2 = 14\) and \(S_1 = 13\).
\(a_2 = S_2 - S_1 = 14 - 13 = 1\)
Step 4: Find the sum of the first 3 terms (\(S_3\)).
Using the formula \(S_n = n + 12\), substitute \(n=3\):
\(S_3 = 3 + 12 = 15\)
Step 5: Find the third term (\(a_3\)).
Using the relationship \(a_n = S_n - S_{n-1}\), for \(n=3\), we have \(a_3 = S_3 - S_2\). We already calculated \(S_3 = 15\) and \(S_2 = 14\).
\(a_3 = S_3 - S_2 = 15 - 14 = 1\)
So, the third term of the series is 1.
Summary of Terms and Sums
| \(n\) | \(S_n = n + 12\) | \(a_n = S_n - S_{n-1}\) (for \(n>1\)) | Term Value (\(a_n\)) |
|---|---|---|---|
| 1 | \(S_1 = 1 + 12 = 13\) | \(a_1 = S_1\) | 13 |
| 2 | \(S_2 = 2 + 12 = 14\) | \(a_2 = S_2 - S_1 = 14 - 13\) | 1 |
| 3 | \(S_3 = 3 + 12 = 15\) | \(a_3 = S_3 - S_2 = 15 - 14\) | 1 |
As shown in the table, the third term (\(a_3\)) is 1.
Revision Table: Key Concepts
| Concept | Description | Formula |
|---|---|---|
| Sum of first \(n\) terms | The total value obtained by adding the first \(n\) terms of a series. | \(S_n\) (given as \(n+12\)) |
| \(n\)-th term (\(a_n\)) | The value of the term at position \(n\) in the series. | \(a_n = S_n - S_{n-1}\) for \(n > 1\), and \(a_1 = S_1\) |
| Finding a specific term | To find the \(k\)-th term, calculate \(S_k\) and \(S_{k-1}\) and subtract. | \(a_k = S_k - S_{k-1}\) |
Additional Information: Types of Series
A series is the sum of the terms of a sequence. The problem deals with finding terms of a series given its sum formula. This series might belong to a specific type, like an arithmetic progression or a geometric progression, but the method used here (\(a_n = S_n - S_{n-1}\)) works for any series, regardless of its type, as long as the formula for \(S_n\) is known.
- Arithmetic Series: A series where the difference between consecutive terms is constant (called the common difference). The sum formula for an arithmetic series is typically a quadratic expression in \(n\), like \(S_n = An^2 + Bn\). In our case, \(S_n = n+12\), which is linear in \(n\). Let's check if it is an arithmetic series. The terms are \(a_1 = 13\), \(a_2 = 1\), \(a_3 = 1\). The difference between \(a_2\) and \(a_1\) is \(1 - 13 = -12\). The difference between \(a_3\) and \(a_2\) is \(1 - 1 = 0\). Since the differences are not constant, this series is not an arithmetic series.
- Geometric Series: A series where the ratio between consecutive terms is constant (called the common ratio). The sum formula involves powers of the common ratio. This series is clearly not geometric either based on the terms we found.
The method \(a_n = S_n - S_{n-1}\) is a universal approach to find the \(n\)-th term from the sum of the first \(n\) terms.
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