One Subtraction Decides Whether a List of Numbers Has a Pattern You Can Use
Test a list of numbers for a constant difference, write the nth term of an arithmetic progression, find the missing first term, common difference or position, decide whether a given number belongs to a sequence, and count a term from the end.
How can you find the hundredth term of a list without writing out the first ninety-nine?
Look at the list and the pattern is obvious: add each time. But "add 3 each time" is a set of instructions, not an answer. Following it to reach the hundredth term means doing ninety-nine additions, and a single slip anywhere ruins the result.
A formula replaces all of that with one multiplication. Because the hundredth term is the first term plus lots of , it is
One line, and no chance of drift. That is what this chapter is about: turning a repeated instruction into a single expression you can evaluate at any position.
A list with a constant difference between consecutive terms is called an arithmetic progression, usually shortened to AP. Two numbers describe it completely:
- **, the first term
- , the common difference, which may be positive, negative or zero
Everything in this part follows from those two letters.** You will test a list for a constant difference, write the general term, work backwards to find , or the position from partial information, decide whether a particular number appears in a sequence at all, and count a term from the last one rather than the first.
One warning at the outset. The difference must be constant between every pair of consecutive terms, not just the first pair. has differences — the gaps themselves keep changing, so it is not an AP, however regular it looks.
This page covers the first part of the CBSE Class 10 Maths chapter on arithmetic progressions: recognising an AP, the nth term formula, finding unknowns from given terms, and terms counted from the end.
A formula replaces all of that with one multiplication. Because the hundredth term is the first term plus lots of , it is
One line, and no chance of drift. That is what this chapter is about: turning a repeated instruction into a single expression you can evaluate at any position.
A list with a constant difference between consecutive terms is called an arithmetic progression, usually shortened to AP. Two numbers describe it completely:
- **, the first term
- , the common difference, which may be positive, negative or zero
Everything in this part follows from those two letters.** You will test a list for a constant difference, write the general term, work backwards to find , or the position from partial information, decide whether a particular number appears in a sequence at all, and count a term from the last one rather than the first.
One warning at the outset. The difference must be constant between every pair of consecutive terms, not just the first pair. has differences — the gaps themselves keep changing, so it is not an AP, however regular it looks.
This page covers the first part of the CBSE Class 10 Maths chapter on arithmetic progressions: recognising an AP, the nth term formula, finding unknowns from given terms, and terms counted from the end.
How do you test whether a list of numbers is an arithmetic progression?
**Subtract each term from the one that follows it. If every difference is the same number, the list is an AP, and that number is the common difference .**
Always subtract the earlier term from the later one, in that order. Reversing it flips the sign of and every later answer with it.
Worked example 1. Is an AP?
The differences are not equal, so it is not an AP. Each term here is double the one before, which is a different kind of pattern entirely.
Worked example 2. Is an AP?
**Yes — it is an AP with and . An irrational common difference is perfectly allowed; the requirement is that it be constant, not that it be a whole number.
Worked example 3.** Is an AP?
**Yes, with and .** The terms increase because is positive, even though they begin negative.
Worked example 4 — a decreasing AP. Is an AP?
**Yes, with and . A negative common difference means the terms decrease, and it is the case where sign errors do the most damage.
Worked example 5 — using the test algebraically.** For what value of do , and form an AP?
If they are in AP, the two differences must be equal:
Check by substituting: the three terms become , then , then . The differences are and . Equal, so the answer is right.
A shortcut worth knowing for three terms in AP. If , , are in AP then , which rearranges to
so the middle term is the average of its neighbours. That single relation answers most three-term questions in one line, and it is why is called the arithmetic mean of and .
The most common error in this section is subtracting the wrong way round. In the common difference is , not . A student who writes will produce an increasing sequence that has nothing to do with the question, and the error will not announce itself until the final answer looks strange.
Always subtract the earlier term from the later one, in that order. Reversing it flips the sign of and every later answer with it.
Worked example 1. Is an AP?
The differences are not equal, so it is not an AP. Each term here is double the one before, which is a different kind of pattern entirely.
Worked example 2. Is an AP?
**Yes — it is an AP with and . An irrational common difference is perfectly allowed; the requirement is that it be constant, not that it be a whole number.
Worked example 3.** Is an AP?
**Yes, with and .** The terms increase because is positive, even though they begin negative.
Worked example 4 — a decreasing AP. Is an AP?
**Yes, with and . A negative common difference means the terms decrease, and it is the case where sign errors do the most damage.
Worked example 5 — using the test algebraically.** For what value of do , and form an AP?
If they are in AP, the two differences must be equal:
Check by substituting: the three terms become , then , then . The differences are and . Equal, so the answer is right.
A shortcut worth knowing for three terms in AP. If , , are in AP then , which rearranges to
so the middle term is the average of its neighbours. That single relation answers most three-term questions in one line, and it is why is called the arithmetic mean of and .
The most common error in this section is subtracting the wrong way round. In the common difference is , not . A student who writes will produce an increasing sequence that has nothing to do with the question, and the error will not announce itself until the final answer looks strange.
Formula
What is the formula for the nth term of an AP?
The first term plus one fewer lot of the common difference than the position number.
**Why and not .** To reach the first term you add zero times; to reach the second you add it once; to reach the tenth you add it nine times. The number of steps is always one less than the number of terms, and that off-by-one is the single most important thing in the chapter.
Worked example 1. For the AP with and , find the general term and the tenth term.
Check by listing: . The tenth term is . Correct, and notice that the simplified general term is a **linear expression in — always true for an AP, and a useful way to recognise one.
Worked example 2 — a decreasing AP.** Find the twentieth term of .
Here and , so
Check with the general term: , and . Correct.
Worked example 3 — writing the next three terms. Write the next three terms of the AP .
The difference is , so the next three terms are
Check with the formula: . Correct, so the listing and the formula agree.
Worked example 4 — a general term that is given. The nth term of a sequence is . Show that it is an AP and find its first term and common difference.
The difference is every time. More neatly, for any ,
**A constant independent of , so it is an AP with and .** Notice that turned out to be the coefficient of — which is always the case when the general term is linear in .
**Why and not .** To reach the first term you add zero times; to reach the second you add it once; to reach the tenth you add it nine times. The number of steps is always one less than the number of terms, and that off-by-one is the single most important thing in the chapter.
Worked example 1. For the AP with and , find the general term and the tenth term.
Check by listing: . The tenth term is . Correct, and notice that the simplified general term is a **linear expression in — always true for an AP, and a useful way to recognise one.
Worked example 2 — a decreasing AP.** Find the twentieth term of .
Here and , so
Check with the general term: , and . Correct.
Worked example 3 — writing the next three terms. Write the next three terms of the AP .
The difference is , so the next three terms are
Check with the formula: . Correct, so the listing and the formula agree.
Worked example 4 — a general term that is given. The nth term of a sequence is . Show that it is an AP and find its first term and common difference.
The difference is every time. More neatly, for any ,
**A constant independent of , so it is an AP with and .** Notice that turned out to be the coefficient of — which is always the case when the general term is linear in .
How do you find n, a or d when only part of the information is given?
**Write , put in everything you know, and solve for the one thing you do not. If two unknowns remain, you need two equations.
Worked example 1 — finding the position.** Which term of the AP is ?
Here , , and :
**So is the thirty-fifth term.
Check**: . Correct.
Worked example 2 — deciding whether a number is a term at all. Is a term of the AP ?
Here and . Suppose is the nth term:
This is not a whole number, and a position must be a positive integer. **So is not a term of this AP.
That is the whole method, and the reasoning is the marked part.** Set the number equal to , solve for , and then ask whether the you get is a positive integer. If it is not, the number does not appear. The neighbouring terms confirm it: and , so falls in the gap.
Worked example 3 — counting how many terms there are. How many two-digit numbers are divisible by ?
The two-digit multiples of are , an AP with , and last term :
**There are such numbers.
Notice the hidden in that answer.** The gap from to holds steps of , but there are numbers, because you must count the one you started from. Every counting question in this chapter turns on that single addition.
Worked example 4 — two unknowns from two given terms. The third term of an AP is and its seventh term is . Find the AP.
Subtracting the first from the second:
and then , so . **The AP is .
Check**: and . Both given conditions hold.
A shortcut that saves a step. Because and ,
so in the example above, gives immediately. **The two positions differ by , so the terms differ by ** — and this relation removes from the working entirely.
Worked example 5 — a proof rather than a number. If the ninth term of an AP is zero, show that its twenty-ninth term is twice its nineteenth term.
From we get , so . Then
**So , as required.** No numerical values were needed, and cancels out of the comparison — which is the mark of a well-posed proof question.
Worked example 1 — finding the position.** Which term of the AP is ?
Here , , and :
**So is the thirty-fifth term.
Check**: . Correct.
Worked example 2 — deciding whether a number is a term at all. Is a term of the AP ?
Here and . Suppose is the nth term:
This is not a whole number, and a position must be a positive integer. **So is not a term of this AP.
That is the whole method, and the reasoning is the marked part.** Set the number equal to , solve for , and then ask whether the you get is a positive integer. If it is not, the number does not appear. The neighbouring terms confirm it: and , so falls in the gap.
Worked example 3 — counting how many terms there are. How many two-digit numbers are divisible by ?
The two-digit multiples of are , an AP with , and last term :
**There are such numbers.
Notice the hidden in that answer.** The gap from to holds steps of , but there are numbers, because you must count the one you started from. Every counting question in this chapter turns on that single addition.
Worked example 4 — two unknowns from two given terms. The third term of an AP is and its seventh term is . Find the AP.
Subtracting the first from the second:
and then , so . **The AP is .
Check**: and . Both given conditions hold.
A shortcut that saves a step. Because and ,
so in the example above, gives immediately. **The two positions differ by , so the terms differ by ** — and this relation removes from the working entirely.
Worked example 5 — a proof rather than a number. If the ninth term of an AP is zero, show that its twenty-ninth term is twice its nineteenth term.
From we get , so . Then
**So , as required.** No numerical values were needed, and cancels out of the comparison — which is the mark of a well-posed proof question.
How do you find a term counted from the end of a finite AP?
Treat the last term as a new first term and walk backwards, so the common difference changes sign. For a finite AP with last term ,
Why the minus. Going forwards you add ; going backwards you subtract it. **The formula is the same with in place of and in place of , so there is nothing extra to memorise.
Worked example 1.** Find the eleventh term from the last term of the AP .
Here and :
**The eleventh term from the end is .
Check by the long route.** First find how many terms there are:
The eleventh from the end is the th term from the start:
The two routes agree, which is the strongest evidence that the shortcut was applied correctly.
Worked example 2. Find the twentieth term from the last term of the AP .
Here and :
Check by counting the terms: gives , so . The twentieth from the end is the nd term:
Correct again.
The relation worth writing down. If an AP has terms in all, then
**The is there for the same reason as before** — both the first and the last term are being counted. Checking the relation on a small case settles it: in a list of terms, the st from the end should be the th from the start, and . It works.
Which method should you use? If the question gives the last term, the shortcut is faster and involves no counting. **If the question also asks how many terms there are, find anyway and use the position relation, since you will have done the work already.
One boundary case to be careful about. The formula needs the actual last term**, not an approximate end of the list. In the value really is a term, as the check confirmed. If a stated final value is not a term of the AP, the question is faulty — and testing it takes the same one line as worked example 2 of the previous section.
Why the minus. Going forwards you add ; going backwards you subtract it. **The formula is the same with in place of and in place of , so there is nothing extra to memorise.
Worked example 1.** Find the eleventh term from the last term of the AP .
Here and :
**The eleventh term from the end is .
Check by the long route.** First find how many terms there are:
The eleventh from the end is the th term from the start:
The two routes agree, which is the strongest evidence that the shortcut was applied correctly.
Worked example 2. Find the twentieth term from the last term of the AP .
Here and :
Check by counting the terms: gives , so . The twentieth from the end is the nd term:
Correct again.
The relation worth writing down. If an AP has terms in all, then
**The is there for the same reason as before** — both the first and the last term are being counted. Checking the relation on a small case settles it: in a list of terms, the st from the end should be the th from the start, and . It works.
Which method should you use? If the question gives the last term, the shortcut is faster and involves no counting. **If the question also asks how many terms there are, find anyway and use the position relation, since you will have done the work already.
One boundary case to be careful about. The formula needs the actual last term**, not an approximate end of the list. In the value really is a term, as the check confirmed. If a stated final value is not a term of the AP, the question is faulty — and testing it takes the same one line as worked example 2 of the previous section.
Exam tip
Which habits keep an AP answer clean?
**Write and on their own line before using any formula.** Half the marks in this chapter are lost by carrying a wrong sign for through an otherwise perfect calculation.
- **Compute as later term minus earlier term**, so has
- Check the difference on at least two consecutive pairs, not one — three equal gaps make a much safer claim than one
- **Use , never .** The tenth term needs nine steps of
- **Simplify into the form ** when you need several terms; the coefficient of is
- **When you solve for , check it is a positive whole number.** A fractional means the number is not a term
- **Add when counting terms between two limits** — to in steps of gives numbers, not
- **For two given terms, use ** to get without finding
- **For a term from the end, use **, or convert the position with
- State whether the AP is increasing or decreasing as a sanity check on the sign of
- A common difference may be negative, fractional or irrational — only constancy is required
The misconception to name. A sequence that looks regular is not automatically an AP. ** doubles**, so the differences keep growing and no single exists; is a list of squares with differences . The test is on the differences, not on how neat the list looks.
A second trap. Treating the position as something that can be a fraction. ** counts terms, so it must be a positive integer**, and the whole method for "is this number a term?" rests on that. A student who reports " is the th term" has done the algebra and missed the question.
- **Compute as later term minus earlier term**, so has
- Check the difference on at least two consecutive pairs, not one — three equal gaps make a much safer claim than one
- **Use , never .** The tenth term needs nine steps of
- **Simplify into the form ** when you need several terms; the coefficient of is
- **When you solve for , check it is a positive whole number.** A fractional means the number is not a term
- **Add when counting terms between two limits** — to in steps of gives numbers, not
- **For two given terms, use ** to get without finding
- **For a term from the end, use **, or convert the position with
- State whether the AP is increasing or decreasing as a sanity check on the sign of
- A common difference may be negative, fractional or irrational — only constancy is required
The misconception to name. A sequence that looks regular is not automatically an AP. ** doubles**, so the differences keep growing and no single exists; is a list of squares with differences . The test is on the differences, not on how neat the list looks.
A second trap. Treating the position as something that can be a fraction. ** counts terms, so it must be a positive integer**, and the whole method for "is this number a term?" rests on that. A student who reports " is the th term" has done the algebra and missed the question.
Did you know
Why does the nth term of an AP look exactly like a straight line?
Simplify the nth term formula and watch what appears:
That is the equation of a straight line, with where usually sits. The common difference is the slope, and is what the line would give at .
So if you plot the terms of an AP against their positions — , , and so on — every point lies exactly on one straight line. For the points , , , sit on the line , rising for every step of .
Which explains several things at once.
- **Why is constant — a straight line has one slope everywhere
- Why the general term is always linear in **, so a sequence given as is an AP and one given as is not
- Why an AP is determined by any two of its terms — two points determine a line
- Why the differences of a non-AP grow or shrink — those sequences lie on curves, not lines
The third point is worth dwelling on. Two terms of an AP, at any two positions, pin down the whole sequence. That is exactly why the third-term-is-, seventh-term-is- question has a unique answer: two points, one line, one AP. And the relation is nothing but the slope formula, rise over run, rearranged.
It also gives a picture for the term from the end. Reading the line from right to left instead of left to right reverses the slope, which is precisely why the backwards formula uses . Same line, opposite direction of travel.
And the plot is genuinely discrete, which is the boundary case to keep straight. The line is continuous and has a value at ; the sequence does not, because there is no two-and-a-half-th term. The line is a scaffold that the terms sit on, not the sequence itself — and that is exactly why coming out as means a number is absent from the list rather than sitting between two terms.
That is the equation of a straight line, with where usually sits. The common difference is the slope, and is what the line would give at .
So if you plot the terms of an AP against their positions — , , and so on — every point lies exactly on one straight line. For the points , , , sit on the line , rising for every step of .
Which explains several things at once.
- **Why is constant — a straight line has one slope everywhere
- Why the general term is always linear in **, so a sequence given as is an AP and one given as is not
- Why an AP is determined by any two of its terms — two points determine a line
- Why the differences of a non-AP grow or shrink — those sequences lie on curves, not lines
The third point is worth dwelling on. Two terms of an AP, at any two positions, pin down the whole sequence. That is exactly why the third-term-is-, seventh-term-is- question has a unique answer: two points, one line, one AP. And the relation is nothing but the slope formula, rise over run, rearranged.
It also gives a picture for the term from the end. Reading the line from right to left instead of left to right reverses the slope, which is precisely why the backwards formula uses . Same line, opposite direction of travel.
And the plot is genuinely discrete, which is the boundary case to keep straight. The line is continuous and has a value at ; the sequence does not, because there is no two-and-a-half-th term. The line is a scaffold that the terms sit on, not the sequence itself — and that is exactly why coming out as means a number is absent from the list rather than sitting between two terms.
Exam relevance
How do arithmetic progressions prepare you for JEE?
This is foundation work for Class 11 Sequences and Series, a recurring JEE Main topic and a regular source of JEE Advanced problems.
Where the nth term leads. Class 11 keeps unchanged and places it beside the geometric progression, , and the harmonic progression. **The habit of identifying and first transfers directly to identifying and , and questions that mix the two — a sequence whose terms are in AP while their logarithms are in GP, for instance — are a JEE Advanced favourite.
Where the arithmetic mean leads.** The relation becomes the general arithmetic mean of numbers, and Class 11 compares it with the geometric mean through the inequality that the arithmetic mean of positive numbers is never less than their geometric mean. That inequality is one of the most used tools in JEE Advanced algebra, and it grows out of the three-term relation you meet here.
Where the insertion questions lead. "Insert arithmetic means between two numbers" is a standard Class 11 item, solved by treating the two given numbers as the first and last terms of an AP with terms. **The counting-with-the- discipline is exactly what it needs.
Where the linear-in- observation leads.** Class 11 identifies a progression from its general term, and a term that is linear in signals an AP while a quadratic signals that the sums are in AP instead. JEE Main sets that identification as a one-liner, and it is answered by the same reasoning as the straight-line picture above.
Where the two-terms-determine-the-AP idea leads. It underlies problems where two conditions on distant terms must be combined, and it is the same reasoning that makes two points determine a line in Straight Lines.
Question types to expect. At this level: test for an AP, find a term, find a position, count terms, and find a term from the end. In competitive papers: mixed AP and GP conditions, arithmetic means inserted between numbers, the arithmetic-mean inequality, and identification of a progression from its general term.
The single trap that costs marks. The off-by-one in . **The tenth term involves nine steps, and there are two-digit multiples of rather than — and the identical error at JEE level appears when counting how many terms of a series fall below a given bound.
A second trap.** Carrying with the wrong sign. In a decreasing AP the common difference is negative, and a single sign slip changes every subsequent term. The cheapest guard is to state "the AP is decreasing, so " before computing anything.
Board versus competitive emphasis. The CBSE paper marks the values of and , the formula, the substitution and the conclusion; a competitive paper marks the final number or the count. **The transferable habit is writing the general term in the simplified form ** — it makes visible as the coefficient, makes any term a one-step substitution, and makes the AP-or-not question answerable at a glance.
Where the nth term leads. Class 11 keeps unchanged and places it beside the geometric progression, , and the harmonic progression. **The habit of identifying and first transfers directly to identifying and , and questions that mix the two — a sequence whose terms are in AP while their logarithms are in GP, for instance — are a JEE Advanced favourite.
Where the arithmetic mean leads.** The relation becomes the general arithmetic mean of numbers, and Class 11 compares it with the geometric mean through the inequality that the arithmetic mean of positive numbers is never less than their geometric mean. That inequality is one of the most used tools in JEE Advanced algebra, and it grows out of the three-term relation you meet here.
Where the insertion questions lead. "Insert arithmetic means between two numbers" is a standard Class 11 item, solved by treating the two given numbers as the first and last terms of an AP with terms. **The counting-with-the- discipline is exactly what it needs.
Where the linear-in- observation leads.** Class 11 identifies a progression from its general term, and a term that is linear in signals an AP while a quadratic signals that the sums are in AP instead. JEE Main sets that identification as a one-liner, and it is answered by the same reasoning as the straight-line picture above.
Where the two-terms-determine-the-AP idea leads. It underlies problems where two conditions on distant terms must be combined, and it is the same reasoning that makes two points determine a line in Straight Lines.
Question types to expect. At this level: test for an AP, find a term, find a position, count terms, and find a term from the end. In competitive papers: mixed AP and GP conditions, arithmetic means inserted between numbers, the arithmetic-mean inequality, and identification of a progression from its general term.
The single trap that costs marks. The off-by-one in . **The tenth term involves nine steps, and there are two-digit multiples of rather than — and the identical error at JEE level appears when counting how many terms of a series fall below a given bound.
A second trap.** Carrying with the wrong sign. In a decreasing AP the common difference is negative, and a single sign slip changes every subsequent term. The cheapest guard is to state "the AP is decreasing, so " before computing anything.
Board versus competitive emphasis. The CBSE paper marks the values of and , the formula, the substitution and the conclusion; a competitive paper marks the final number or the count. **The transferable habit is writing the general term in the simplified form ** — it makes visible as the coefficient, makes any term a one-step substitution, and makes the AP-or-not question answerable at a glance.
Key takeaways
What must you be able to do from this part?
One test, one formula and one careful count.
- An AP has a constant difference between consecutive terms, and later term earlier term
- Check at least two pairs. is not an AP, because its differences are
- ** may be negative, fractional or irrational** — is an AP with
- The nth term is , and the number of steps is always one fewer than the position
- **With and **, and ; **with and **,
- **The general term of an AP is linear in **, and the coefficient of is
- **Three terms in AP satisfy — the middle term is the arithmetic mean of its neighbours
- To find a position**, solve ; ** is the th term of
- A number is a term only if comes out a positive integer** — is not a term of because
- **Counting needs the **: there are two-digit multiples of , from to
- Two given terms fix the AP, and gives without finding
- The nth term from the end of a finite AP is , which equals the th term from the start
- **The eleventh term from the end of is **, and the twentieth from the end of is
- Plotting terms against positions gives a straight line of slope , but only at whole-number positions
The quickest self-test is a two-way check. Take the AP , find its fifteenth term from the start and its eleventh from the end, and see whether you get the same number both ways.
- An AP has a constant difference between consecutive terms, and later term earlier term
- Check at least two pairs. is not an AP, because its differences are
- ** may be negative, fractional or irrational** — is an AP with
- The nth term is , and the number of steps is always one fewer than the position
- **With and **, and ; **with and **,
- **The general term of an AP is linear in **, and the coefficient of is
- **Three terms in AP satisfy — the middle term is the arithmetic mean of its neighbours
- To find a position**, solve ; ** is the th term of
- A number is a term only if comes out a positive integer** — is not a term of because
- **Counting needs the **: there are two-digit multiples of , from to
- Two given terms fix the AP, and gives without finding
- The nth term from the end of a finite AP is , which equals the th term from the start
- **The eleventh term from the end of is **, and the twentieth from the end of is
- Plotting terms against positions gives a straight line of slope , but only at whole-number positions
The quickest self-test is a two-way check. Take the AP , find its fifteenth term from the start and its eleventh from the end, and see whether you get the same number both ways.