Quantitative Aptitude · Sequence and Series

Arithmetic Progression (AP) for CA Foundation

Updated 1 October 2026 · Fact-checked

An arithmetic progression is a sequence in which each term differs from the previous one by a fixed number, the common difference d. Find a and d, then use the nth term Tn = a + (n − 1)d or the sum Sn = n/2 × [2a + (n − 1)d]. Solve for n when it is unknown.

Understand Arithmetic Progression (AP)

A sequence is a list of numbers in a fixed order. In an arithmetic progression (AP), you get each term by adding the same number to the term before it. That number is the common difference, written d.

For example, 5, 8, 11, 14 is an AP with first term a = 5 and d = 3. If d is positive, the AP rises. If d is negative, it falls. If d = 0, every term is the same.

To test whether a list is an AP, subtract each term from the next. If every difference is equal, it is an AP. Check all gaps, not just the first two.

The nth term is the first term plus (n − 1) steps of d. That is why the formula has (n − 1), not n. The sum of n terms uses a pairing idea: the first and last terms add to the same total as the second and second-last. So the sum is the number of terms times the average of the first and last term.

In the exam, most AP questions ask you to find a term, the number of terms, a sum, or to set up terms from conditions given in words.

Key formulas to remember

Common difference: d = T2 − T1 = Tn − T(n−1); Must be the same for every consecutive pair.
nth term: Tn = a + (n − 1)d; a is the first term. Use it to find any term or the number of terms.
Sum of n terms (using d): Sn = n/2 × [2a + (n − 1)d]; Use when a, d and n are known.
Sum of n terms (using last term): Sn = n/2 × (a + l); l is the last term, l = Tn. Use when the last term is known.
Term from sum: Tn = Sn − S(n−1); Valid for n ≥ 2. T1 = S1.
Number of terms: n = (l − a) ÷ d + 1; Rearranged nth term formula. n must be a positive whole number.
Three terms in AP: If a, b, c are in AP, then 2b = a + c; The middle term is the average of its neighbours.
Selecting terms: 3 terms: a − d, a, a + d. 4 terms: a − 3d, a − d, a + d, a + 3d. 5 terms: a − 2d, a − d, a, a + d, a + 2d; Symmetric choice makes the sum easy. For 4 terms the common difference is 2d.
Sum of first n natural numbers: 1 + 2 + … + n = n(n + 1) ÷ 2; AP with a = 1, d = 1.

How to solve Arithmetic Progression (AP) questions

Use this method for any AP question, whether it asks for a term, a sum, the number of terms or a word problem.

1Confirm it is an AP by checking that the consecutive differences are equal.
2Write down what is given: a, d, n, l, Tn or Sn. Mark clearly what is asked.
3Turn any words into equations. For example, 'the 5th term is 17' becomes a + 4d = 17.
4If a and d are both unknown, form two equations and solve them as simultaneous equations.
5Pick the formula: Tn for a term or n, Sn for a total. Use Sn = n/2 × (a + l) if you know the last term.
6Substitute carefully, with brackets around negative values of d.
7For n, check it is a positive whole number. If not, recheck the data or the arithmetic.
8Match your answer with the options and check it by plugging back in if time allows.

Quickest way: Shortcuts and option elimination for AP MCQs

When to use it: Use under time pressure in the 2-hour objective paper, where each wrong answer costs 0.25 marks.

Remember that the term-to-term jump is d. To go from the pth term to the qth term, add (q − p)d. This saves finding a.
For the sum, use average × count. Sn = n × (average of first and last term).
If the number of terms is odd, the sum equals n × middle term.
For selecting terms, use a − d, a, a + d. The sum of three terms is 3a, so you get a at once.
If a sum is 0 or terms are symmetric, check whether the middle term is 0.
Use options: test an option in the nth term formula rather than solving fully when the algebra is long.
Check the unit digit or parity of your answer against the options to remove two choices quickly.
Skip questions needing three or more long steps with unclear data, and return later.

Common mistakes in Arithmetic Progression (AP)

Using n instead of (n − 1) in Tn = a + (n − 1)d.
Students remember 'nd' from the pattern and forget the first term is already counted.
Fix: Test the formula on n = 1. You must get T1 = a. If not, the formula is wrong.
Finding d by subtracting in the wrong order, giving the wrong sign.
Taking the earlier term minus the later term in a decreasing AP.
Fix: Always compute later term − earlier term. For 20, 17, 14, d = 17 − 20 = −3.
Getting n = (l − a) ÷ d and forgetting to add 1.
Students count the gaps instead of the terms.
Fix: Number of terms = number of gaps + 1. Check with 3, 5, 7: gaps = 2, terms = 3.
Using Sn = n/2 × (a + l) when l is not the nth term, or with the wrong n.
Mixing up the last term with a term somewhere else in the AP.
Fix: Make sure l is the final term of the n terms you are adding. If unsure, use n/2 × [2a + (n − 1)d].
Choosing four terms as a − d, a, a + d, a + 2d and then assuming equal gaps of d around a.
Extending the three-term pattern without care.
Fix: For four terms use a − 3d, a − d, a + d, a + 3d, where the common difference is 2d.
Assuming Tn = Sn − S(n−1) works for n = 1.
Applying the formula blindly when S0 is not defined.
Fix: Use T1 = S1 and apply the difference formula only from n = 2.

Worked examples

Example 1

The 4th term of an AP is 14 and the 9th term is 34. What is the sum of its first 10 terms? (A) 170 (B) 180 (C) 190 (D) 200

Show the solution

T4 = a + 3d = 14 and T9 = a + 8d = 34.
Subtract: 5d = 20, so d = 4.
Then a + 12 = 14, so a = 2.
S10 = 10/2 × [2(2) + 9(4)] = 5 × (4 + 36) = 5 × 40 = 200.
Check: T10 = 2 + 36 = 38. S10 = 10/2 × (2 + 38) = 200.

Answer: (D) 200

Example 2

How many terms are there in the AP 7, 13, 19, …, 151? (A) 23 (B) 24 (C) 25 (D) 26

Show the solution

a = 7, d = 13 − 7 = 6, l = 151.
n = (l − a) ÷ d + 1 = (151 − 7) ÷ 6 + 1.
144 ÷ 6 = 24.
n = 24 + 1 = 25.
Check: T25 = 7 + 24 × 6 = 7 + 144 = 151.

Answer: (C) 25

Example 3

Three numbers in AP have a sum of 27 and a product of 693. What is the largest of the three numbers? (A) 11 (B) 13 (C) 15 (D) 17

Show the solution

Let the numbers be a − d, a, a + d.
Sum = 3a = 27, so a = 9.
Product = 9(81 − d²) = 693, so 81 − d² = 77.
d² = 4, so d = 2 or −2.
The numbers are 7, 9, 11 in either order.
Check: 7 + 9 + 11 = 27 and 7 × 9 × 11 = 693.
The largest is 11.

Answer: (A) 11

Exam tips

Look at the options first. If the data gives an easy a and d, the answer often comes in one line.
Questions on 'sum of the first n natural, odd or even numbers' are standard AP sums. Use the shortcut: odd numbers sum to n², even numbers sum to n(n + 1).
For word problems with equal yearly or monthly increases (salary rises, instalments), set up an AP and decide whether you need a term or a sum.
If n comes out as a fraction or negative, you have made an error. Recheck before marking.
Do not guess when two options remain after elimination unless you are fairly sure, because a wrong answer costs 0.25 marks.

Practice questions from Sequence and Series

Arithmetic Progression (AP): frequently asked questions

How do I find the number of terms in an AP?

Use n = (l − a) ÷ d + 1, where l is the last term. Make sure n is a positive whole number. If it is not, the given last term is not part of the AP.

What is the difference between Tn and Sn?

Tn is the single nth term of the AP. Sn is the total of the first n terms. Tn = a + (n − 1)d and Sn = n/2 × [2a + (n − 1)d].

How do I know if a sequence is an AP?

Subtract each term from the next one. If all differences are equal, it is an AP. Check every pair, because two equal gaps are not enough proof.

How do I choose terms when the sum or product is given?

For 3 terms use a − d, a, a + d. For 4 terms use a − 3d, a − d, a + d, a + 3d. The d terms cancel in the sum, so you can find a quickly.