The sum of the first n terms of an arithmetic sequence is 153, with first term 5 and common difference 3. Find n. - GetMeFoodie
How to Solve for n When the Sum of an Arithmetic Sequence Is 153
How to Solve for n When the Sum of an Arithmetic Sequence Is 153
Ever stumbled across a math puzzle that stirred quiet fascination? The sum of the first n terms of an arithmetic sequence hitting 153—starting at 5 with a common difference of 3—might seem like a quiet equation, yet it’s quietly resonating in digital spaces. Curious minds are naturally drawn to puzzles that connect logic, curiosity, and real-world patterns. This article breaks down how to decode the value of n behind this sum—without sensationalism, preserving clarity, accuracy, and relevance to today’s US audience.
The sum of the first n terms of an arithmetic sequence is 153, with first term 5 and common difference 3. Find n.
Understanding the Context
Why This Calculation Is Gaining Attention
In recent years, linear learning—particularly through educational apps and social platforms—has sparked curiosity in how simple math models real-life scenarios. The arithmetic sequence offers a clean, predictable pattern: start at 5, rise by 3 each step. The idea that such a structure fits a total of 153 invites deeper exploration. Users searching online aren’t news-seekers—they’re learners, students, educators, and professionals using math to understand trends, budgets, or growth models. This problem reflects a broader trend: people seeking clarity in numbers behind everyday experiences. Platforms that deliver clear, factual explanations earn trust and visibility, especially on mobile where curious users scroll quickly but linger on trusted insights.
How It Actually Works: A Clear Explanation
An arithmetic sequence follows a fixed pattern: each term increases by a constant difference—in this case, 3. The general formula for the sum of the first n terms is:
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Key Insights
Sₙ = n/2 × (2a + (n−1)d)
Where:
Sₙ = sum of the first n terms
a = first term = 5
d = common difference = 3
n = number of terms (what we’re solving for)
Plugging in known values:
153 = n/2 × (2×5 + (n−1)×3)
153 = n/2 × (10 + 3n − 3)
153 = n/2 × (3n + 7)
Multiply both sides by 2:
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306 = n(3n + 7)
306 = 3n² + 7n
Rearranging into standard quadratic form:
3n² + 7n — 306 = 0
Apply the quadratic formula: n = [-b ± √(b² − 4ac)] / 2a
With a = 3, b = 7, c = -306:
Discriminant: b² − 4ac = 49 + 3672 = 3721
√3721 = 61
n = [-7 ± 61] / 6
Two solutions:
n = (−7 + 61)/6 = 54/6 = 9
n = (−7 − 61)/6 = −68/6 (discard—n must be positive)
Therefore, n = 9 is the solution.
Verification: Sum of first 9 terms with a = 5, d = 3:
S₉ = 9/2 × (2×5 + 8×3) = 4.5 × (10 + 24) = 4.5 × 34 = 153 — confirms accuracy.
