The sum of an arithmetic sequence is exactly 1150. The first term is 10, and the common difference increases by 5 each term. After how many terms is the total sum 1150? - RTA
Why Are More People Solving “Sum of an Arithmetic Sequence Is Exactly 1150 With First Term 10 and Increasing Difference?”
The sum of an arithmetic sequence is exactly 1150. The first term is 10, and the common difference increases by 5 each term. After how many terms is the total sum 1150?
Why Are More People Solving “Sum of an Arithmetic Sequence Is Exactly 1150 With First Term 10 and Increasing Difference?”
The sum of an arithmetic sequence is exactly 1150. The first term is 10, and the common difference increases by 5 each term. After how many terms is the total sum 1150?
In a digital landscape growing more analytical and curious by the day, a growing number of users are exploring patterns hidden in everyday math—like how a sequence grows without a clear pattern. This particular problem draws attention because it blends foundational math with a subtle twist: the common difference isn’t constant, making it more engaging than a standard arithmetic problem. People are naturally curious when numbers tell a story shaped by change, not just steady steps—mirroring real-world trends where growth shifts unpredictably.
This sequence begins at 10, and each next term increases by 5 more than the last: 10, then 10 + (5), then 10 + (5+5), etc. The challenge lies in determining exactly how many such adjusted steps are needed so the cumulative total reaches 1150.
Understanding the Context
How It Actually Works
To solve for the number of terms when the total sum equals 1150, start by defining the sequence formally:
- First term: $ a = 10 $
- Difference increases by 5 each term: $ d_1 = 5, d_2 = 10, d_3 = 15, \ldots $
The $ n $-th term of this evolving sequence is:
$ a_n = 10 + \sum_{k=1}^{n-1} (5k) $
This summation reflects the increasing gap between terms.
The overall sum of the sequence up to $ n $ terms, $ S_n $, is therefore:
$ S_n = 10n + \sum_{k=1}^{n-1} \frac{5k(k+1)}{2} $
The factor $ \frac{5k(k+1)}{2} $ comes from the sum of the cumulative differences.
Simplifying step-by-step:
The sum of the first $ n-1 $ multiples of 5’s weighted average yields a quadratic expression in $ n $. Plugging into the formula for the sum of an arithmetic series, the final combined sum becomes:
$ S_n = 10n + \frac{5}{2}(
