An arithmetic sequence is a fundamental concept in algebra where the difference between consecutive terms remains constant. That said, problems often present the n-th term or the sum of the series while asking for the very beginning: the first term. This fixed value, known as the common difference, allows us to predict any term in the progression if we possess enough information. Mastering how to find the first term of an arithmetic sequence requires a solid grasp of the explicit formula and the ability to manipulate algebraic equations based on the given data points.
Understanding the Core Formula
Before diving into specific scenarios, You really need to establish the primary tool for this task. The explicit formula for the n-th term of an arithmetic sequence is:
$a_n = a_1 + (n - 1)d$
Where:
- $a_n$ represents the value of the n-th term (the term you know). Worth adding: * $a_1$ represents the first term (the unknown you are solving for). So * $n$ represents the position of the known term (the term number). * $d$ represents the common difference (the amount added or subtracted to get to the next term).
To find the first term ($a_1$), you simply rearrange this formula to isolate $a_1$:
$a_1 = a_n - (n - 1)d$
This rearrangement is the key to almost every problem variation you will encounter. The challenge lies in identifying the values for $a_n$, $n$, and $d$ from the problem statement.
Scenario 1: Given a Specific Term and the Common Difference
This is the most direct application of the formula. Here's the thing — g. The problem explicitly provides a term later in the sequence (e., "the 10th term is 50") and the common difference (e.g., "the common difference is 3").
Example: Find the first term of an arithmetic sequence where the 12th term ($a_{12}$) is 73 and the common difference ($d$) is 5 Small thing, real impact..
Step 1: Identify the known variables.
- $a_n = a_{12} = 73$
- $n = 12$
- $d = 5$
Step 2: Substitute into the rearranged formula. $a_1 = 73 - (12 - 1)(5)$
Step 3: Solve using order of operations. $a_1 = 73 - (11)(5)$ $a_1 = 73 - 55$ $a_1 = 18$
Verification: Start at 18 and add 5 eleven times: 18, 23, 28, 33, 38, 43, 48, 53, 58, 63, 68, 73. The 12th term is indeed 73 Easy to understand, harder to ignore..
Scenario 2: Given Two Specific Terms (Common Difference Unknown)
Often, a problem will not give you the common difference directly. Here's the thing — instead, it provides two terms at different positions, such as "the 5th term is 20 and the 12th term is 41. " In this case, you must calculate $d$ first.
The Logic: The difference between the two known term values equals the common difference multiplied by the number of steps (gaps) between their positions.
$d = \frac{a_y - a_x}{y - x}$
Example: The 4th term ($a_4$) is 15 and the 9th term ($a_9$) is 35. Find the first term Easy to understand, harder to ignore..
Step 1: Calculate the common difference ($d$). There are $9 - 4 = 5$ steps between the 4th and 9th term. The value increased by $35 - 15 = 20$. $d = \frac{20}{5} = 4$
Step 2: Use one known term to find $a_1$. Let's use the 4th term ($a_4 = 15$, $n=4$, $d=4$). $a_1 = a_4 - (4 - 1)d$ $a_1 = 15 - (3)(4)$ $a_1 = 15 - 12$ $a_1 = 3$
Alternative Check: Use the 9th term ($a_9 = 35$, $n=9$, $d=4$). $a_1 = 35 - (9 - 1)(4)$ $a_1 = 35 - 32 = 3$ Both methods yield the same result, confirming the answer.
Scenario 3: Given the Sum of the First n Terms ($S_n$)
Sometimes the problem provides the sum of the first $n$ terms ($S_n$) instead of a specific term value. You will need the sum formula:
$S_n = \frac{n}{2}(2a_1 + (n - 1)d)$ Alternatively: $S_n = \frac{n}{2}(a_1 + a_n)$
If you are given $S_n$, $n$, and $d$, use the first version. If you are given $S_n$, $n$, and $a_n$ (the last term), use the second version.
Example A (Given $S_n$, $n$, and $d$): The sum of the first 10 terms ($S_{10}$) is 235. The common difference ($d$) is 3. Find the first term.
Step 1: Plug knowns into the sum formula. $235 = \frac{10}{2}(2a_1 + (10 - 1)3)$
Step 2: Simplify and solve for $a_1$. $235 = 5(2a_1 + 27)$ $47 = 2a_1 + 27$ $20 = 2a_1$ $a_1 = 10$
Example B (Given $S_n$, $n$, and $a_n$): The sum of the first 8 terms is 100. The 8th term ($a_8$) is 23. Find the first term Nothing fancy..
Step 1: Use the alternative sum formula. $S_8 = \frac{8}{2}(a_1 + a_8)$ $100 = 4(a_1 + 23)$
Step 2: Solve for $a_1$. $25 = a_1 + 23$ $a_1 = 2$
Scenario 4: Given Two Sums ($S_n$ and $S_m$)
This is a more advanced variation often found in competitive exams or higher-level coursework. You might be given the sum of the first 5 terms ($S_5$) and the sum of the first 10 terms ($S_{10}$), asked to find $a_1$.
Strategy: Set up a system of two equations with two unknowns ($a_1$ and $d$) using the sum formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$.
Example: $S_5 = 40$ and $S_{10} = 115$. Find $a_1$.
Equation 1 (for $n=5$): $40 = \frac{5}{2}(2a_1 + 4d)$ Multiply by 2: $80 = 5(2a_
Equation 1 (for (n=5)):
[ 40=\frac{5}{2}\bigl(2a_1+4d\bigr) ]
Multiply both sides by 2:
[ 80 = 5\bigl(2a_1+4d\bigr) ]
Expand:
[ 80 = 10a_1 + 20d ]
Divide by 10 to simplify:
[ 8 = a_1 + 2d \qquad\text{(1)} ]
Equation 2 (for (n=10)):
[ 115 = \frac{10}{2}\bigl(2a_1+9d\bigr)=5\bigl(2a_1+9d\bigr) ]
Expand:
[ 115 = 10a_1 + 45d ]
Divide by 5:
[ 23 = 2a_1 + 9d \qquad\text{(2)} ]
Solving the System
From (1) we have (a_1 = 8 - 2d). Substitute this expression into (2):
[ 2(8-2d) + 9d = 23 ]
[ 16 - 4d + 9d = 23 ]
[ 16 + 5d = 23 ]
[ 5d = 7 ;\Longrightarrow; d = \frac{7}{5}=1.4 ]
Now plug (d) back into (1):
[ a_1 = 8 - 2!\left(\frac{7}{5}\right)=8-\frac{14}{5}= \frac{40-14}{5}= \frac{26}{5}=5.2 ]
Result:
[ \boxed{a_1 = \frac{26}{5};(5.2)\qquad d = \frac{7}{5};(1.4)} ]
A quick verification confirms the sums:
- (S_5 = \frac{5}{2}\bigl(2\cdot5.2 + 4\cdot1.4\bigr)=2.5(10.4+5.6)=2.5(16)=40)
- (S_{10}=5\bigl(2\cdot5.2 + 9\cdot1.4\bigr)=5(10.4+12.6)=5(23)=115)
Both match the given values, so the solution is consistent.
Closing Thoughts
We have explored four common ways to determine the first term of an arithmetic sequence:
- Two explicit terms – compute the common difference from the gap between positions, then back‑solve for (a_1).
- A term and the common difference – directly apply the term formula (a_n =
$a_1 + (n-1)d$ and rearrange.
4. That said, 3. Plus, A sum ($S_n$), the number of terms ($n$), and either the common difference ($d$) or the last term ($a_n$) – substitute into the appropriate sum formula ($S_n = \frac{n}{2}(2a_1 + (n-1)d)$ or $S_n = \frac{n}{2}(a_1 + a_n)$) and solve the resulting linear equation. Two distinct sums ($S_n$ and $S_m$) – construct a system of two equations in $a_1$ and $d$, solve for the common difference first, then back-substitute to find the first term.
This is the bit that actually matters in practice.
Regardless of which scenario you encounter, the workflow remains consistent: identify the known variables, select the formula(s) that link those variables to $a_1$, and solve algebraically. Mastering these four patterns equips you to handle virtually any "find the first term" problem in arithmetic sequences, from standard textbook exercises to competition-level challenges.
