Overview
A sequence is governed by a relationship between positions and values. Large indices can make a problem appear computationally long, but the governing rule may compress the work dramatically. An arithmetic sequence repeats a difference, a geometric sequence repeats a ratio, and a recurrence can repeat a short cycle. Your first task is to recognize which pattern the information actually supports; your second is to keep positions, steps and requested quantities separate.
Count the steps between known terms
An arithmetic sequence has fourth term 17 and ninth term 42. The increase in value is 25, but it occurs over five steps, because 9 − 4 = 5. Therefore the common difference is 25/5 = 5. To recover the first term from the fourth, move back three steps: 17 − 3·5 = 2.
The number of steps between terms is the difference in their indices, not the number of terms you can name including both endpoints. This is why the fourth term is three differences beyond the first. The distractor 5 gives the common difference rather than the first term; 12 and 22 move only one step from 17. Label each number as either a position, a term value or a change. Then verify your starting term by rebuilding both known terms. A method that explains only one given term has not yet used all the evidence.
Inspect a recurrence for a cycle
The recurrence a₁ = 5 and aₙ₊₁ = 3 − aₙ gives the next term −2. Applying the rule again gives 5, returning to the starting value. The sequence therefore alternates 5, −2, 5, −2, and so on. Odd-numbered positions hold 5; even-numbered positions hold −2. Since 202 is even, a₂₀₂ = −2.
Why does observing the return establish a permanent cycle? The same input is passed through the same deterministic rule, so it must give the same next output. Once the sequence returns to 5, its next two steps repeat. You do not need to calculate 202 terms or guess from the size of the index. Still, the starting index matters: the first term is indexed by 1, not 0. Marking the first two positions prevents a reversed parity pattern, which is the reason the answer 5 remains a plausible distractor.
Separate a geometric term from a partial sum
For the sequence 81, 27, 9, …, each term is one third of the previous term. The next two terms are 3 and 1. The first five terms therefore sum to 81 + 27 + 9 + 3 + 1 = 121. With only five terms, writing them explicitly is clear and efficient; a memorized sum formula is optional rather than necessary.
A common ratio acts between consecutive terms, not between consecutive partial sums. The total after four terms is 120; adding the fifth gives 121. This makes 120 a particularly close wrong answer: it reflects a correct calculation stopped one term too early. As a separate illustration, counting the last term twice would give 122. When asked for a sum, mark each term as you include it and check that the number of included terms matches the requested index. A term question and a sum question can share the same sequence while asking for very different objects.
Use structure before calculation
These three examples call for different shortcuts because their rules differ. The arithmetic problem uses a constant change per index step. The recurrence problem uses a repeated state. The geometric problem uses a constant multiplier and then adds a finite list. Trying to force all three into one formula can conceal the very feature that makes each manageable.
For a new sequence, read whether the rule is given explicitly or must be inferred from enough stated information. An initial list alone may support many possible patterns, so rely on descriptions such as “arithmetic” or “geometric” when the problem supplies them. Before practice, explain aloud what changes and what stays fixed at each step. At the end, check the requested index against your counting convention and distinguish the requested term from an intermediate difference, ratio or total. This habit is particularly valuable when an intimidating index is meant to distract from a short, repeating structure.
Apply the method to a new case
A reservoir initially contains 10 liters. On day n, for n = 1 through 4, its net gain is 3n − 1 liters. Find the volume after day 4. The formula describes daily changes, not the total volume: the gains are 2, 5, 8 and 11 liters. Their sum is 26, which must be added to the initial 10, giving 36 liters. Reporting 11 confuses the fourth change with the final state; reporting 26 omits the initial condition. Check by updating the state successively: 12, 17, 25 and 36 liters. The constant increase of 3 between gains makes the gains an arithmetic sequence, but the reservoir totals are not arithmetic. Recognize what the index labels before selecting a sequence formula, and use the units and the initial state to distinguish a term from an accumulated sum.