Arithmetic progressions are a Class 10 chapter and appear often in aptitude tests. Enter the first term, common difference and number of terms to get the nth term, the sum and the first few terms.
Enter the first term and common difference.
Enter how many terms you need.
Read the nth term and the sum.
nth term: aₙ = a + (n − 1)d
Sum: Sₙ = n ÷ 2 × (2a + (n − 1)d)
An arithmetic progression, or AP, is a list of numbers that goes up or down by the same amount each time, such as 5, 8, 11, 14. This calculator takes the first term a, the common difference d and the number of terms n, and gives the nth term, the sum of the first n terms and the opening few terms so you can see the pattern.
APs form a full chapter in Class 10 NCERT mathematics and appear in almost every aptitude test. They also describe many everyday patterns: a savings plan where you put aside ₹500 more each month, seats in an auditorium that increase by a fixed number per row, instalments that fall by a fixed amount, or the positions of evenly spaced fence posts. When n is large, adding term by term is tedious, and the sum formula does it in one step.
1. Identify the first term a and the common difference d = (second term − first term). Check that the difference is the same throughout.
2. Decide how many terms n you need.
3. Find the nth term: aₙ = a + (n − 1)d.
4. Find the sum of the first n terms: Sₙ = n ÷ 2 × (2a + (n − 1)d).
5. If you already know the last term l, use the shorter form Sₙ = n ÷ 2 × (a + l).
6. To find how many terms reach a given value, solve a + (n − 1)d = value for n and check it is a whole number.
The first term already sits at a with no difference added. The second term adds d once, the third adds it twice, and so the nth term adds d exactly n − 1 times. That gives aₙ = a + (n − 1)d. Plotted against n, the terms lie on a straight line with slope d, which is why APs are the discrete version of linear functions. A positive d gives an increasing sequence, a negative d a decreasing one, and d = 0 a constant list.
Write the sum forwards and then backwards underneath it. Each column adds the first and last terms, then the second and second-last, and every such pair has the same total, a + l. There are n columns, so twice the sum is n(a + l), and Sₙ = n(a + l) ÷ 2. Replacing l by a + (n − 1)d gives Sₙ = n ÷ 2 × (2a + (n − 1)d). The legend says the young Gauss used this to add 1 to 100 in moments: 50 pairs of 101 make 5,050.
Simple interest produces an AP: each year adds the same fixed amount, so the balance after n years follows a + (n − 1)d. Compound interest produces a geometric progression instead, where each term is multiplied by a fixed ratio. Over short periods the two can look similar, but a GP eventually outgrows any AP. The sum of an AP grows roughly with n², because it is the area of a triangle-like staircase, while a single term grows only linearly with n.
Sanya starts a savings habit by keeping aside ₹5,000 in the first month and ₹500 more than the previous month each month after that, for 24 months.
nth term: aₙ = a + (n − 1)d a24 = 5000 + (24 − 1) × 500 = 16,500
Sum of n terms: Sₙ = n ÷ 2 × (2a + (n − 1)d) S24 = 24 ÷ 2 × (2 × 5000 + 23 × 500) = 2,58,000
Answer: Sum of first 24 terms 2,58,000; 24th term 16,500; First terms 5,000, 5,500, 6,000, 6,500, 7,000, 7,500…
Using n instead of n − 1 in the nth term, which overshoots by one difference.
Finding d by subtracting in the wrong order, which gives the wrong sign for a decreasing sequence.
Treating a sequence with a constant ratio, such as 3, 6, 12, 24, as an AP.
Confusing the nth term with the sum of n terms in word problems.
Getting a fractional n when asked which term equals a value, and not realising that value is not in the sequence.
Planning savings or deposits that increase by a fixed amount each month.
Counting seats in rows that each have a fixed number more than the previous row.
Tracking simple interest balances year by year.
Solving Class 10 board questions and aptitude problems on sequences.
Estimating total distance when speed increases by a fixed step each lap or day.
Can the common difference be negative?
Yes, then the sequence decreases, for example 20, 17, 14.
How do I find d?
Subtract any term from the next one: d = a₂ − a₁.