Algebra · Eqora guide
Arithmetic sequence homework: count the gaps
Find a term without moving one step too far. Separate positions, differences and sums, then check one troublesome step with Eqora.
Your arithmetic sequence homework gives 7, 11, 15, … and asks for the twelfth term. You correctly notice an increase of four, then write 7 + 12 × 4 = 55. The arithmetic is fine, but the answer is one term too far along. The first term is already present before you make a single move. Reaching the twelfth position therefore needs eleven changes, not twelve. Counting terms and counting gaps are different jobs.
We will build the formula from that distinction, use it with increasing and decreasing sequences, and separate a term’s value from its position and from a sum. Each example assumes an arithmetic sequence with indices beginning at one unless stated otherwise. After explaining the method on paper, we will use Eqora to inspect the precise step that caused confusion and then solve a changed problem without copying a response.
Eqora publishes this guide and is a learning aid, not a substitute for exam work or a guarantee of correctness or grades. Check the original question, notation, assumptions and result yourself. The snowy garden photographs suggest separate objects and the spaces between them. They are conceptual illustrations, not scale drawings or diagrams whose stone counts or distances encode the numerical examples. Your written calculation remains the evidence.
Arithmetic sequence homework starts with a constant change
An arithmetic sequence has the same difference between every pair of consecutive terms. Subtract the earlier value from the next one: 11 − 7 = 4 and 15 − 11 = 4. We call this signed difference d. OpenStax’s arithmetic sequences lesson distinguishes a constant difference from a constant ratio and explains both explicit and recursive descriptions. Adding four repeatedly is not the same operation as multiplying every term by four.
A short list alone does not determine a unique infinite sequence. Many rules could start with the same three numbers and change later. Use the arithmetic formula when the exercise states that the sequence is arithmetic, supplies a constant-change rule, or asks you to work under that assumption. Checking the displayed differences helps detect a transcription error; it does not prove a claim about every undisplayed term without the given model.
Twelve terms contain eleven forward moves
Start at position one. Position two needs one addition of d, position three needs two, and position four needs three. The number of moves is always one less than the destination position. This gives aₙ = a₁ + (n − 1)d for positive integer n. The parentheses matter: the whole count n − 1 multiplies d. Writing a₁ + n − 1d changes the grouping and does not express the same reasoning.
For our example, a₁₂ = 7 + (12 − 1) × 4 = 7 + 44 = 51. A quick independent check lists nearby values: a₁₀ = 43, a₁₁ = 47 and a₁₂ = 51. The rejected answer 55 is a₁₃, which reveals the extra move. This is an indexing error rather than a multiplication error. Naming the type of mistake helps you choose a useful correction instead of recalculating the same incorrect setup more carefully.
Test the formula at n = 1 before using a distant index. It gives 7 + 0 × 4 = 7, exactly the starting value. The faulty expression 7 + 4n instead gives eleven at position one. You can also simplify the correct expression to aₙ = 4n + 3; substituting one still gives seven. A simplified formula may look different while keeping the same starting condition and the same four-unit change between neighboring positions.
Negative and zero differences follow the same rule
Consider 24, 21, 18, … with a₁ = 24. The ordered subtraction is 21 − 24 = −3, so d = −3. The ninth term is a₉ = 24 + (9 − 1)(−3) = 24 − 24 = 0. Eight downward moves take you from the first position to the ninth. A decreasing sequence does not need a different formula; it needs a negative difference and careful parentheses around that negative number.
Indexing can also begin at zero, especially in computing. If the initial term is called a₀, the appropriate expression is aₙ = a₀ + nd. There are n moves from position zero to position n. This is not a contradiction of the earlier formula: the starting label changed. Read the definition instead of automatically importing the index convention from another exercise. If the sequence is finite, also check that the requested position exists.
Two distant terms reveal the difference per gap
Suppose the exercise gives a₄ = 19 and a₁₀ = 43 for an arithmetic sequence. The total change is 43 − 19 = 24, but it occurs across 10 − 4 = 6 gaps. Therefore d = 24/6 = 4. Dividing by two because two values are printed would count pieces of information, not moves. Dividing by ten would start the journey at the wrong position. The index difference identifies the actual number of equal changes.
Recover the first term from the fourth: a₁ = 19 − (4 − 1) × 4 = 7. Verify the other given value independently: 7 + (10 − 1) × 4 = 43. Both endpoints must fit the reconstructed sequence. If they do not, check the index subtraction, the sign of the value subtraction and the copied numbers. Do not round a fractional difference merely to make the terms look more familiar; arithmetic sequences need not have integer values.
A recursive description would say a₁ = 7 and aₙ = aₙ₋₁ + 4 for n ≥ 2. That rule describes one move at a time, while the explicit formula jumps directly to a chosen position. Both need an initial value. Our guide to explaining solutions in your own words helps you connect a symbolic step to a sentence: six gaps each contributing four create a total change of twenty-four, regardless of how many terms were printed.
A value belongs only if its position is allowed
To ask whether 51 occurs in 7, 11, 15, …, solve 51 = 7 + (n − 1)4. Subtract seven to get 44 = 4(n − 1), divide by four to get eleven, then add one: n = 12. This positive integer is an allowed position in the infinite sequence. Substitute it into the original expression to verify the value, rather than treating a rearranged equation as an automatically reliable answer.
For 52, the same process gives n = 12.25. That is a solution of the linear equation considered over real numbers, but it is not a term position in this sequence. A sequence does not contain a quarter of a position between twelve and thirteen. If the task lists only the first ten terms, even n = 12 would fall outside its range. The domain restriction belongs to the answer, not to an optional note after the calculation.
When d is nonzero, n = 1 + (value − a₁)/d is a convenient rearrangement. When d = 0, do not divide by it. Check whether the proposed value equals the repeated term: if it does, every allowed position works; otherwise none does. These checks distinguish an algebraic solution from a meaningful response to the original task. Our independent answer-checking guide develops the same habit of checking assumptions and admissible values.
The last term is not the sum of all terms
Our twelfth term is 51, but the sum of the first twelve terms is a different quantity. Write S₁₂ when you mean 7 + 11 + … + 51. Pair the first and last values: 7 + 51 = 58. The second and next-to-last also add to 58 because one rises by four while the other falls by four. There are six such pairs, so the sum is 6 × 58 = 348.
In general, Sₙ = n(a₁ + aₙ)/2 for an arithmetic sequence. The expression averages the endpoints and multiplies that average by the number of terms. For twelve terms, 12(7 + 51)/2 = 348. This formula counts n terms, unlike the n − 1 changes used to find aₙ. Do not transfer the gap count into the sum formula. The two counts serve different purposes, even though they refer to the same sequence.
Read whether the question asks for a term, its position, or a total before selecting an operation. In a savings model, a monthly deposit and the total of several deposits are not interchangeable. Keep this interpretation separate from any real-world interest or fees, which our simple arithmetic model does not include. A correct numeric result attached to the wrong quantity is still an incorrect response to that particular question.
Use Eqora to investigate the extra move
First try the problem yourself and mark the uncertain line, such as 7 + 12 × 4. If you use Eqora, capture the whole readable question, including the requested index and any statement that the sequence is arithmetic. Avoid glare, cropped subscripts or a page containing unrelated exercises. Our photo guide for math problems explains how to keep the original notation legible. The photograph shown here uses a blank notebook as an illustration, not a real submitted task.
Check the captured or entered task against the original before evaluating an explanation. Distinguish a₁ from a₀, twelve from two, and a minus sign from a dash. Then inspect the steps: identify the initial term, verify the ordered subtraction for d, and test the expression at n = 1. A response giving 51 is not enough by itself. You want to understand why the eleven moves occur and whether the explanation respects this exercise’s assumptions.
Ask a focused follow-up: Why does reaching the twelfth term require eleven additions? Show the check at position one without solving my next practice problem. If the explanation changes the given numbers or skips the indexing issue, return to the written task and inspect that mismatch. Eqora can support this conversation, but you still need to judge the reasoning. Follow your course’s rules for permitted assistance and keep assessment work your own.
Inspect the reasoning, then change the example
Before moving on, explain the distinction aloud: the first value is already counted, so twelve positions require eleven transitions. Point to the starting index in your own notes. You should be able to reject the extra-move formula by substituting n = 1 without needing an answer key. If you cannot yet give that explanation, ask about that one step again rather than requesting another complete worked solution to memorize.
Now put the response aside and try 5, 8, 11, … as an arithmetic sequence beginning at one. Find the fifteenth term, decide whether fifty appears, and compute the sum of the first fifteen terms. Set up all three calculations before reading the check. Here d = 3, so a₁₅ = 5 + 14 × 3 = 47. Solving 50 = 5 + (n − 1)3 gives n = 16, an allowed position in the infinite sequence.
For the sum, S₁₅ = 15(5 + 47)/2 = 390. You can check the average independently: the middle, eighth term is 26, and fifteen terms average to that value in this symmetric arithmetic list. Compare each result with the exact quantity asked for. If your term was fifty, you probably took one extra move again. Correct the count, explain the correction, and finish with a new example whose starting value or difference you choose yourself.
Keep the check when the numbers change
Record the correction: twelve positions need eleven changes because the initial term occupies position one. Add the formula and a substitution checking the start. On the next homework page, read the model and index convention afresh. Work on paper and use assistance to examine a specific uncertainty, not to replace your attempt.
