Arithmetic Progressions

How is the common difference identified?

Subtract each term from the next, keeping the order consistent. For the original example 32, 27, 22, 17, the difference is −5. Reversing the subtraction produces the wrong sign. A sequence with repeated equal terms is also an arithmetic progression: 6, 6, 6 has difference zero. These cases matter because an arithmetic progression need not increase.

When only a few terms are shown, the problem must state or justify that the pattern continues. Four observed values alone cannot prove an infinite sequence follows one rule. In an examination exercise that explicitly calls the sequence an AP, use that condition. In a real data series, do not infer a permanent arithmetic pattern merely from a short run of equal differences.

Why is the nth term a plus n minus one differences?

Call the first term a and the common difference d. The second term is a + d, the third a + 2d, and the fourth a + 3d. Reaching position n requires n − 1 moves from the first position. Therefore a_n = a + (n − 1)d. This derivation is a useful defence against the common off-by-one error a + nd.

For an original seating example, the first row has 14 seats and each following row has 3 more. The twelfth row has 14 + 11 × 3 = 47 seats. The result concerns the twelfth row alone. It does not answer how many seats all twelve rows contain; that is a sum question with a different operation.

How can two terms determine the sequence?

Suppose the fourth term is 19 and the ninth is 44. There are five equal steps between those positions, so 5d = 44 − 19 = 25 and d = 5. Since the fourth term is a + 3d, the first term is 19 − 15 = 4. The sequence begins 4, 9, 14, 19.

Write positional differences explicitly. The ninth and fourth positions are five steps apart, not six. A general relation is a_m − a_k = (m − k)d. It allows a common difference to be recovered without first knowing the first term. Substitute the recovered values into both given terms as a quick independent check.

How is the sum obtained?

For n terms, pair the first with the last, the second with the second-last, and so on. Each pair has the same total a + l, where l is the last term. Averaging the endpoints therefore gives the average of the terms. Multiplying by their number gives S_n = n(a + l)/2. Replacing l with a + (n − 1)d yields S_n = n[2a + (n − 1)d]/2.

In the seating example, twelve rows total 12(14 + 47)/2 = 366 seats. This also works for an odd number of terms: the middle term equals the endpoint average and is counted once. The formula does not require physically forming whole pairs when n is odd.

How is membership of a number tested?

Set the proposed value equal to a + (n − 1)d and solve for n. In the AP 7, 11, 15, …, the value 83 gives 83 = 7 + 4(n − 1), so n = 20. It is the twentieth term. The value 84 would give n = 20.25, which is not a valid positive integer position.

A calculated position must also lie within the stated finite sequence. If the sequence contains only ten terms, a value occurring at position twenty is outside it. For d = 0, every term equals a, so membership must be checked directly rather than by dividing by d. These boundary checks prevent algebraically neat but contextually impossible answers.

How do counting problems become AP problems?

Multiples of 7 between 20 and 80 are 21, 28, …, 77. The first and last admissible values are determined before counting. The number of terms is (77 − 21)/7 + 1 = 9. The plus one counts both endpoints. Their sum is 9(21 + 77)/2 = 441.

If the question says strictly between two multiples, exclude those boundary values before applying the formula. Keep the interval condition separate from the AP calculation. Errors often occur when a candidate counts the numerical gap correctly but forgets whether the first and last admissible terms are included.

What distinguishes fixed increments from fixed percentages?

An invented deposit plan adds ₹40 more than the previous deposit each month: ₹100, ₹140, ₹180. The deposits form an AP. A balance growing by 10 percent each period follows ₹100, ₹110, ₹121, whose differences are 10 and 11. Equal percentages are applied to changing bases and therefore do not generally give an AP.

Also distinguish periodic deposits from the accumulated balance. Even when deposits form an AP, their running total follows the sum formula and usually has changing first differences. Identify what the sequence measures before using any formula. A reliable solution names the quantity, finds the step, checks the index and then chooses term or sum.

How can the term formula be checked?

Substitute n = 1: the formula must return the first term because no common-difference step has yet occurred. Substitute n = 2: exactly one step should appear. These boundary checks expose the frequent error of using n differences instead of n minus one differences without requiring a long expansion of the sequence.