Applications of APs

A stack of logs with one fewer log in each higher layer, a saver who adds a little more to their piggy bank every month, a hall whose rows of seats grow by a fixed number each time you move back — none of these look like "arithmetic progressions" at first glance, yet each one is exactly that: a sequence where a quantity changes by the same fixed amount at every step. This is the payoff topic of the AP chapter. All the machinery built earlier — spotting the first term $a$ , the common difference $d$ , the $n$ th-term rule, and the sum formula — now gets aimed at real, worded situations. The real skill tested here is not new algebra; it is translation: reading a paragraph of English, recognising the AP hiding inside it, and choosing the right formula to answer the actual question asked.

The core insight

Any real-life situation where a quantity increases or decreases by a fixed amount at each step is an AP in disguise — translate it into $a$ , $d$ , $n$ and then pick $a_n$ for "a specific stage" or $S_n$ for "a running total."

The core idea

Every application problem on APs follows the same four-step translation, no matter how different the story sounds.

Spot the AP. Read past the story to find the one quantity that changes by a constant amount each step — rupees saved per month, seats per row, bricks per layer, metres drilled per day. Only that quantity forms the AP; a word problem often mentions several numbers, and only one of them is the sequence.

Fix $a$ and $d$ . The very first value in the pattern is $a$ ; the constant step size is $d$ , taken as positive when the pattern grows and negative when it shrinks.

Match the question to a formula. If the question names a single stage — "how much in the 8th instalment," "how many seats in row 12" — it wants the $n$ th term:

a_n = a + (n-1)d

where: $a_n$ is the value at stage $n$ , $a$ is the first-stage value, and $d$ is the fixed step. If the question instead asks for a cumulative or total amount — "total saved after 20 months," "total seats in the hall" — it wants the sum of the first $n$ terms:

S_n = \frac{n}{2}\big[2a + (n-1)d\big]

where: $S_n$ is the total of the first $n$ terms, and $a$ , $d$ , $n$ carry the same meaning as before. Choosing between these two formulas correctly is the single most exam-relevant decision in this topic.

Solve, then interpret. Substitute the known values, solve the resulting linear or (often) quadratic equation for the unknown, and translate the number back into the story's language — discarding any root that cannot represent a real count of months, rows, or days. The syllabus keeps this to simple, direct situations only: steady savings or repayments, seating that grows by a constant row-to-row amount, stacked objects losing or gaining a fixed count per layer, and steady production or distance patterns. No multi-stage or compound scenarios are expected at this level.

Figure

A row of five stacked bars increasing in height by equal steps illustrating an arithmetic progression used to model a real situation

Derivation ADeriving the sum formula used in every AP application

The goal is to see why $S_n = \tfrac{n}{2}[2a+(n-1)d]$ is true, since almost every "total after n steps" application problem in this topic rests on this one formula, and knowing where it comes from makes it far easier to trust and remember under exam pressure.

Step 1. Write the sum of the first $n$ terms both forwards and backwards. Forwards: $S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d]$ . Backwards, the same sum can be listed starting from the last term: $S_n = [a+(n-1)d] + [a+(n-2)d] + \cdots + a$ .

Step 2. Add the two lines term by term. Every pair of matching positions adds up to the same value, $2a + (n-1)d$ , because whatever is subtracted from one side is added back on the other. Since there are $n$ such pairs, adding the two equations gives $2S_n = n\big[2a + (n-1)d\big]$ .

Step 3. Divide both sides by 2 to isolate $S_n$ , giving $S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]$ . This is exactly the formula used in every application problem that asks for a "total" — a total saved, a total seating capacity, a total distance covered — so an application problem is really just this formula wearing a costume. Sanity check: for $a=1$ , $d=1$ , $n=100$ (the classic Gauss sum $1+2+\cdots+100$ ), the formula gives $S_{100} = \tfrac{100}{2}[2(1)+99(1)] = 50 \times 101 = 5050$ , matching the well-known result.

Worked example

Question. A person saves ₹32 in the first month, ₹36 in the second month, ₹40 in the third month, and so on, increasing the saving by a fixed amount every month. (a) In how many months will their total savings reach ₹2000? (b) How much do they save in that final month?

Step 1. Identify the AP. The monthly savings are $32, 36, 40, \ldots$ , so $a = 32$ and $d = 36 - 32 = 4$ .

Step 2. Recognise this as a total, so use $S_n$ . We want $S_n = 2000$ :

2000 = \frac{n}{2}\big[2(32) + (n-1)(4)\big]

2000 = \frac{n}{2}(4n+60) = n(2n+30)

2n^2 + 30n - 2000 = 0 \quad\Rightarrow\quad n^2 + 15n - 1000 = 0.

Step 3. Factorise: the numbers $40$ and $-25$ multiply to $-1000$ and add to $15$ , so $(n+40)(n-25) = 0$ , giving $n = 25$ or $n = -40$ . Since $n$ must be a positive whole number of months, $n = 25$ .

Step 4. Answer part (b) using $a_n = a+(n-1)d$ : the 25th-month saving is $a_{25} = 32 + 24(4) = 32 + 96 = 128$ .

Answer. The savings reach ₹2000 in the 25th month, and that month's own saving is ₹128.

Competitive stretch

The same auditorium-seating pattern from the syllabus — first row 20 seats, each next row 2 more, 15 rows total — gives $S_{15} = \tfrac{15}{2}[2(20)+14(2)] = \tfrac{15}{2}(68) = 510$ seats, confirming that the identical $S_n$ formula answers seating-capacity questions just as it answered the savings question above.

Common traps

Misreading which quantity is the AP. A story may mention several numbers; only the one changing by a fixed amount each step is the actual AP.
Confusing "a specific stage" with "the total so far." "How much in the 8th month" needs $a_n$ ; "total after 8 months" needs $S_n$ — swapping them gives a completely wrong type of answer.
Keeping an invalid root. Word problems on APs often lead to a quadratic in $n$ ; always discard negative or non-integer solutions since $n$ counts real months, rows, or days.
Off-by-one slips. Check that "the first" case really is $n=1$ (so $a_1 = a$ ) — miscounting the starting stage throws off every later term.

Ask the tutor: If you're stuck deciding whether a word problem wants a single term or a running total, describe the situation to the AI tutor and it will help you set up the right equation step by step.

How do I know whether a word problem needs the nth-term formula or the sum formula?

Look for the exact wording. Phrases like "in the 8th month," "in row 12," or "on the 20th day" ask for one specific stage — use $a_n = a+(n-1)d$ . Phrases like "total saved," "total seating capacity," or "how many in all after n days" ask for a running total — use $S_n = \tfrac{n}{2}[2a+(n-1)d]$ . Reading the question twice before writing any formula avoids most mistakes here.

My word problem gives a quadratic equation with two roots — which one do I keep?

Keep the root that makes sense as a count of real things — months, rows, days, or layers, which must be a positive whole number. If one root is negative or a fraction, discard it without hesitation; it is a mathematical solution to the equation but not a physical answer to the story. If both roots look valid, re-check the problem statement, since application problems at this level are set up to have exactly one sensible answer.

Why does the same sum formula work for savings, seats, and stacked logs — they all seem so different?

Because underneath the different stories, all three situations are the same mathematical object: a sequence where each term is the previous one plus a fixed amount $d$ . Whether that fixed amount is rupees, seats, or logs is just labelling — the $S_n$ formula only cares about $a$ , $d$ , and $n$ , not what they represent. Once you translate the story into these three numbers, the formula does not need to know the context at all.

Applications of APs

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads