Chapter 6.3 — Exercise 6.2 — nth Term of AP

The Formula

From the general form a, a+d, a+2d, a+3d, …, the pattern is that the nth term always adds d exactly (n−1) times to the first term:

When an AP has a fixed final term, that last term is written l instead of aₙ, but it's calculated exactly the same way. Every quantity on the right-hand side — a, n, and d — has to be known before aₙ can be found; the moment any one of the four letters in the formula is the unknown instead, the same equation still applies, just rearranged to solve for whichever piece is missing.

Five Blanks, One Formula

Given any three of a, d, n, and aₙ, the formula solves for whichever one is missing:

The last row confirms something already established in the previous exercise: when d = 0, every term equals the first term, no matter how far out you go — the 105th term is still just 3.5.

Finding a Specific Term, Both Directions

For AP 10, 7, 4, … (a=10, d=−3), the 30th term is a₃₀ = 10 + 29(−3) = −77. Running the formula in reverse answers a different kind of question: for AP 3, 8, 13, 18, … (a=3, d=5), which term equals 78?

So 78 is the 16th term. The same reverse-solving approach checks whether a number belongs to an AP at all: for 11, 8, 5, 2, … (a=11, d=−3), testing whether −150 is a term gives n = 164/3 — not a whole number, so −150 is never a term of this AP, however far you counted.

Working Backward From Two Known Terms

Knowing any two terms of an AP is enough to reconstruct the whole sequence, since two terms pin down both a and d.

The same idea scales up: given that the 11th term of an AP is 38 and the 16th term is 73, subtracting gives 5d = 73−38 = 35, so d=7. Substituting back gives a=−32, and the 31st term is a₃₁ = −32 + 30(7) = 178. Whether the two known terms are close together or far apart, the method is identical — subtract to isolate d, then substitute to find a. The number of steps between the two known positions (5, in this case, from the 11th term to the 16th) always becomes the coefficient of d once the two nth-term equations are subtracted, which is why writing out a₁₆−a₁₁ explicitly, rather than jumping straight to an answer, avoids miscounting that gap.

Six More Unknowns, Same Formula

None of the six problems below hand over a complete AP directly. Each one gives just enough scattered information — two terms, a relationship between two terms, or a comparison between two entirely separate APs — for the nth-term formula, applied once or twice, to pin down whatever's actually being asked. Some of these questions run the formula forward, from a known position to an unknown value; others run it backward, from a known value to an unknown position; a couple don't ask about a single term at all, but use the same formula to compare two APs against each other.

• Which term of an AP is 0, if the 3rd term is 4 and the 9th term is −8? Solving gives d=−2, a=8 — and substituting into aₙ=0 shows the 5th term is 0.
• Find d, if the 17th term exceeds the 10th term by 7. Since a₁₇−a₁₀ = 7d by the formula directly, d = 1.
• Two APs share the same common difference; the difference between their 100th terms is 100. What's the difference between their 1000th terms? Since aₙ−bₙ = (a−b) for any n when both APs share d, the gap never changes — still 100.
• For what n are the nth terms equal in APs 63, 65, 67, … and 3, 10, 17, …? Setting 63+2(n−1) = 3+7(n−1) and solving gives n = 13.
• Find the AP whose 3rd term is 16, where the 7th term exceeds the 5th by 12. From a₇−a₅=2d=12, d=6; then a=4. The AP is 4, 10, 16, 22, …
• Find the first three terms of an AP where a₄+a₈=24 and a₆+a₁₀=44. Solving the resulting pair of equations gives d=5, a=−13, so the first three terms are −13, −8, −3.

Counting Special Numbers

Treating a whole category of numbers as an AP turns a counting problem into a formula problem. Three-digit numbers divisible by 7 run 105, 112, 119, …, 994 — an AP with a=105, d=7, and last term 994. Solving 994=105+(n−1)(7) gives n=128 such numbers. Multiples of 4 between 10 and 250 run 12, 16, 20, …, 248 — a=12, d=4, giving n=60 multiples. The general move is the same in both cases: identify the smallest qualifying number as a, the fixed gap between consecutive qualifying numbers as d, the largest qualifying number as l, and let the nth-term formula do the counting that listing every value by hand would make tedious and error-prone.

The 20th Term From the End

For the AP 3, 8, 13, …, 253, counting from the end works the same way as counting from the start — just with the roles of a and l swapped. This AP has 51 terms total, so the 20th term from the end is the (51−20+1) = 32nd term from the start: a₃₂ = 3 + 31(5) = 158.

Subba Rao's Salary

Subba Rao started a job in 1995 on a monthly salary of ₹5000, with a fixed ₹200 increment every year. In which year did his salary reach ₹7000?

His salary reached ₹7000 in the 11th year from 1995 — that is, in 2005. Recognising the situation as an AP is the whole trick here: the starting salary plays the role of a, the fixed yearly increment plays the role of d, and "which year" is simply asking which value of n makes aₙ equal the target salary — a word problem that looks unrelated to a numeric sequence until it's translated into one.

From One Term to All of Them

Every problem here found a single term, or confirmed one didn't exist. Exercise 6.3 asks a different question entirely — not what one term equals, but what an entire run of terms adds up to — and derives a second formula built directly on top of the nth-term formula from this exercise. For the AP definitions this exercise assumes, revisit Exercise 6.1 or the chapter introduction.