Add Every Number to a Hundred in One Line by Folding the List in Half
Learn to test whether a sequence is an arithmetic progression, find its nth term and locate a given value, derive the sum of the first n natural numbers by pairing, and apply the rule to real growing patterns.
How do you add every number from 1 to 100 without adding them?
Write the numbers to in a row. Underneath, write the same numbers backwards.
Add the two rows column by column. Every column gives the same total:
There are columns, each worth , so the two rows together come to . But the two rows are the same sum written twice, so the sum itself is half of that:
One multiplication and one division, in place of ninety-nine additions.
The trick works because the numbers go up in equal steps, so pairing the ends always gives the same total. A list with equal steps is called an arithmetic progression, and this page covers the second part of the CBSE Class 9 Mathematics chapter on sequences — recognising an AP, finding its terms, and the pairing argument that produces the sum formula.
Add the two rows column by column. Every column gives the same total:
There are columns, each worth , so the two rows together come to . But the two rows are the same sum written twice, so the sum itself is half of that:
One multiplication and one division, in place of ninety-nine additions.
The trick works because the numbers go up in equal steps, so pairing the ends always gives the same total. A list with equal steps is called an arithmetic progression, and this page covers the second part of the CBSE Class 9 Mathematics chapter on sequences — recognising an AP, finding its terms, and the pairing argument that produces the sum formula.
How do you tell whether a sequence is an arithmetic progression?
Subtract each term from the one after it. If every answer is the same, it is an AP, and that shared answer is the common difference .
Worked example 1 — an AP. , , , . Differences: , , . So it is an AP with first term and .
Worked example 2 — not an AP. , , , . Differences: , , . Not constant, so not an AP. These are increasing differences, which the previous part of this chapter identified as the triangular pattern shifted.
Worked example 3 — a decreasing AP. , , , . Differences: each time. So and . A negative common difference is perfectly ordinary, and it is what a falling quantity looks like.
Worked example 4 — fractions. , , , . Differences: each time, so and .
Worked example 5 — not an AP, though it looks regular. , , , . Differences: , , — growing. The ratios are constant at , which makes it a geometric progression, the subject of the next part of this chapter.
Worked example 6 — the natural numbers themselves. , , , is an AP with and . So is every list of consecutive whole numbers, and that is why the pairing trick from the opening applies to them.
**Worked example 7 — finding from two given terms.** In an AP the third term is and the seventh is . Between them lie four steps, so
and working back two steps from the third term, . The AP is
Check every difference, not just the first one. The sequence , , , has differences , , — it passes the first test and fails the second, and a student who checked only would call it an AP and then get every later answer wrong. One extra subtraction rules that out.
Constant difference is exactly what makes the pairing argument work. In the ends pair as and — the same total, because moving in from one end adds while moving in from the other subtracts it. That cancellation is the whole reason the sum formula in the third section exists, and it fails the moment the differences vary.
Worked example 1 — an AP. , , , . Differences: , , . So it is an AP with first term and .
Worked example 2 — not an AP. , , , . Differences: , , . Not constant, so not an AP. These are increasing differences, which the previous part of this chapter identified as the triangular pattern shifted.
Worked example 3 — a decreasing AP. , , , . Differences: each time. So and . A negative common difference is perfectly ordinary, and it is what a falling quantity looks like.
Worked example 4 — fractions. , , , . Differences: each time, so and .
Worked example 5 — not an AP, though it looks regular. , , , . Differences: , , — growing. The ratios are constant at , which makes it a geometric progression, the subject of the next part of this chapter.
Worked example 6 — the natural numbers themselves. , , , is an AP with and . So is every list of consecutive whole numbers, and that is why the pairing trick from the opening applies to them.
**Worked example 7 — finding from two given terms.** In an AP the third term is and the seventh is . Between them lie four steps, so
and working back two steps from the third term, . The AP is
Check every difference, not just the first one. The sequence , , , has differences , , — it passes the first test and fails the second, and a student who checked only would call it an AP and then get every later answer wrong. One extra subtraction rules that out.
Constant difference is exactly what makes the pairing argument work. In the ends pair as and — the same total, because moving in from one end adds while moving in from the other subtracts it. That cancellation is the whole reason the sum formula in the third section exists, and it fails the moment the differences vary.
Formula
How do you find the nth term of an AP and locate a given value?
**Start at the first term and add the common difference times:**
The is the point to be careful about — reaching the fourth term takes three steps, not four.
Worked example 1. For , , , with , :
Check the small cases: and . Both correct.
Worked example 2 — a decreasing AP. For , , with :
Worked example 3 — locating a value. Which term of , , , equals ?
So is the twentieth term, agreeing with worked example 1.
Worked example 4 — another location. Which term of , , , is ? Here , :
Check: . Correct.
Worked example 5 — a value that is not a term. Is a term of , , , ?
But is not divisible by , so would not be a whole number. ** is not a term of this AP — and that is a complete and correct answer, not a failed calculation.
Worked example 6 — the sum of an AP.** The pairing idea gives the sum of any AP, not just the natural numbers. Pair the first and last terms:
For , , , up to the twentieth term :
**A negative or a fractional means the value is not in the list.** The formula is solved as an ordinary equation, but only whole-number solutions count, because there is no two-and-a-half-th term. **Checking that came out a positive whole number is the last step of every location question.**
The is the point to be careful about — reaching the fourth term takes three steps, not four.
Worked example 1. For , , , with , :
Check the small cases: and . Both correct.
Worked example 2 — a decreasing AP. For , , with :
Worked example 3 — locating a value. Which term of , , , equals ?
So is the twentieth term, agreeing with worked example 1.
Worked example 4 — another location. Which term of , , , is ? Here , :
Check: . Correct.
Worked example 5 — a value that is not a term. Is a term of , , , ?
But is not divisible by , so would not be a whole number. ** is not a term of this AP — and that is a complete and correct answer, not a failed calculation.
Worked example 6 — the sum of an AP.** The pairing idea gives the sum of any AP, not just the natural numbers. Pair the first and last terms:
For , , , up to the twentieth term :
**A negative or a fractional means the value is not in the list.** The formula is solved as an ordinary equation, but only whole-number solutions count, because there is no two-and-a-half-th term. **Checking that came out a positive whole number is the last step of every location question.**
Why does the sum of the first n numbers equal n(n+1)/2?
**Because writing the sum forwards and backwards and adding gives identical columns**, each worth .
The derivation. Let be the sum:
Adding column by column, every column totals — the first is , the second is , and each step forward in one row is cancelled by a step back in the other. There are columns, so
Worked example 1 — the hundred. :
Worked example 2. .
Worked example 3. .
Worked example 4 — a sum that does not start at 1. Find . Take the sum to and remove the sum to :
Check another way. There are terms, and pairing the ends gives every time, so
The same answer from two independent routes.
**Worked example 5 — the sum of the first odd numbers.** The odds form an AP with , , so the th odd number is . Pairing:
Check: . Correct — and this is the square-number recursion from the previous part of this chapter, now proved rather than observed.
**Worked example 6 — the sum of the first even numbers.** The evens give , , and the th is :
Check: , and . Correct — exactly twice the sum of the naturals, as it must be.
The formula always gives a whole number, and it has to. Of the two consecutive numbers and , exactly one is even, so the product is always even and dividing by leaves no remainder. A fractional answer from this formula is therefore always an arithmetic slip, never a genuine result.
And these sums are the triangular numbers. is the same expression, so and . The stacked-dots picture and the pairing argument describe one object — which is why the next section's real-world patterns keep producing them.
The derivation. Let be the sum:
Adding column by column, every column totals — the first is , the second is , and each step forward in one row is cancelled by a step back in the other. There are columns, so
Worked example 1 — the hundred. :
Worked example 2. .
Worked example 3. .
Worked example 4 — a sum that does not start at 1. Find . Take the sum to and remove the sum to :
Check another way. There are terms, and pairing the ends gives every time, so
The same answer from two independent routes.
**Worked example 5 — the sum of the first odd numbers.** The odds form an AP with , , so the th odd number is . Pairing:
Check: . Correct — and this is the square-number recursion from the previous part of this chapter, now proved rather than observed.
**Worked example 6 — the sum of the first even numbers.** The evens give , , and the th is :
Check: , and . Correct — exactly twice the sum of the naturals, as it must be.
The formula always gives a whole number, and it has to. Of the two consecutive numbers and , exactly one is even, so the product is always even and dividing by leaves no remainder. A fractional answer from this formula is therefore always an arithmetic slip, never a genuine result.
And these sums are the triangular numbers. is the same expression, so and . The stacked-dots picture and the pairing argument describe one object — which is why the next section's real-world patterns keep producing them.
How do you solve a growing-pattern problem using an AP?
**Find the common difference from the wording, write the th term, then use the sum formula if a total is wanted.
Worked example 1 — chairs at tables.** One square table seats people, and each extra table pushed against the row adds seating for more. The sequence is , , , with , :
Check: , . For twenty tables, seats.
Worked example 2 — a stack of bricks. The bottom row has bricks and each row above has fewer. How many rows reach a top row of bricks, and how many bricks are used?
The sequence is , , , with , . For the last term :
So there are rows. The total is
Check independently. The rows are all the even numbers from to , which is , using . The same answer.
Worked example 3 — a salary rising each year. A starting salary of per month rises by each year. The monthly salary in the eighth year is
Worked example 4 — a triangular arrangement of logs. Logs are stacked with in the bottom row and one fewer in each row above, up to a single log at the top. That is rows, and the total is
Worked example 5 — handshakes. In a room of people, each shakes hands once with every other. Take them one at a time: one shakes hands, the next adds new ones, the next , and so on down to :
For people that is handshakes. Check by counting differently: each of people shakes hands, giving , but every handshake was counted from both ends, so . Two routes, one answer.
Worked example 6 — a savings plan. Someone saves in the first month and more each month than the month before. The total over two years is an AP with , , :
Read whether the question wants a TERM or a TOTAL. Worked example 2 asks for both, and they are bricks in the top row against bricks altogether — two very different numbers from the same setup. **The words in the fifteenth row and in the whole stack are what distinguish them, and mixing them up is the commonest error in these problems.
Check that the pattern really has a constant difference before using any of this.** A stack whose rows go , , is not an AP, and both the th-term formula and the sum formula would give confident wrong answers.
Worked example 1 — chairs at tables.** One square table seats people, and each extra table pushed against the row adds seating for more. The sequence is , , , with , :
Check: , . For twenty tables, seats.
Worked example 2 — a stack of bricks. The bottom row has bricks and each row above has fewer. How many rows reach a top row of bricks, and how many bricks are used?
The sequence is , , , with , . For the last term :
So there are rows. The total is
Check independently. The rows are all the even numbers from to , which is , using . The same answer.
Worked example 3 — a salary rising each year. A starting salary of per month rises by each year. The monthly salary in the eighth year is
Worked example 4 — a triangular arrangement of logs. Logs are stacked with in the bottom row and one fewer in each row above, up to a single log at the top. That is rows, and the total is
Worked example 5 — handshakes. In a room of people, each shakes hands once with every other. Take them one at a time: one shakes hands, the next adds new ones, the next , and so on down to :
For people that is handshakes. Check by counting differently: each of people shakes hands, giving , but every handshake was counted from both ends, so . Two routes, one answer.
Worked example 6 — a savings plan. Someone saves in the first month and more each month than the month before. The total over two years is an AP with , , :
Read whether the question wants a TERM or a TOTAL. Worked example 2 asks for both, and they are bricks in the top row against bricks altogether — two very different numbers from the same setup. **The words in the fifteenth row and in the whole stack are what distinguish them, and mixing them up is the commonest error in these problems.
Check that the pattern really has a constant difference before using any of this.** A stack whose rows go , , is not an AP, and both the th-term formula and the sum formula would give confident wrong answers.
Exam tip
Exam tip: count the steps, not the terms
**The th term is — reaching the fourth term takes three** steps. Writing is the single commonest slip in this chapter.
**Write and on their own line** before substituting anything. For that is , .
Check every difference, not just the first. is not an AP.
**A negative is normal and is what a decreasing pattern looks like.
For "which term equals ", solve as an equation and check is a positive whole number. If it is not, write "not a term of this AP" — that is the full answer.
Two sum formulas, one idea**: when you know the last term, and when you do not.
Show the pairing derivation when asked for it: write the sum forwards and backwards, note columns of , then halve.
****, so to is . **For a sum not starting at , subtract**: .
Learn the two side results: the first odd numbers sum to ; the first even numbers sum to .
Read whether a TERM or a TOTAL is wanted — bricks in the top row against in the stack.
And if the formula returns a fraction, it is a slip: ** is always even**, so the sum is always a whole number.
**Write and on their own line** before substituting anything. For that is , .
Check every difference, not just the first. is not an AP.
**A negative is normal and is what a decreasing pattern looks like.
For "which term equals ", solve as an equation and check is a positive whole number. If it is not, write "not a term of this AP" — that is the full answer.
Two sum formulas, one idea**: when you know the last term, and when you do not.
Show the pairing derivation when asked for it: write the sum forwards and backwards, note columns of , then halve.
****, so to is . **For a sum not starting at , subtract**: .
Learn the two side results: the first odd numbers sum to ; the first even numbers sum to .
Read whether a TERM or a TOTAL is wanted — bricks in the top row against in the stack.
And if the formula returns a fraction, it is a slip: ** is always even**, so the sum is always a whole number.
Did you know
Why every second triangular number is a multiple of the row count
The sum can be computed two slightly different ways depending on whether is odd or even, and the difference is worth seeing.
When is even, the list pairs up perfectly. For : pair with , with , and so on, giving exactly pairs of :
Every number has a partner and nothing is left over.
When is odd, there is a leftover in the middle. For : pair with , with , with , with — four pairs of — and the sits alone:
Check with the formula: . Correct.
And notice what the lonely middle number is. For it is , which is the average of the whole list, and the answer is — the count times the middle value. That works for every odd : to gives , and to gives .
So for an odd-length list the sum is simply how many times the middle one, and no halving is needed at all. For an even-length list there is no middle number — the average falls between two terms, which is exactly why the formula carries a division by .
The same reasoning works on any AP, not just the naturals. The sum is always the number of terms times the average of the first and last, which is what says. The pairing trick and the average rule are one method described from two sides — and for an odd-length list you can see the average sitting in the middle of the row.
When is even, the list pairs up perfectly. For : pair with , with , and so on, giving exactly pairs of :
Every number has a partner and nothing is left over.
When is odd, there is a leftover in the middle. For : pair with , with , with , with — four pairs of — and the sits alone:
Check with the formula: . Correct.
And notice what the lonely middle number is. For it is , which is the average of the whole list, and the answer is — the count times the middle value. That works for every odd : to gives , and to gives .
So for an odd-length list the sum is simply how many times the middle one, and no halving is needed at all. For an even-length list there is no middle number — the average falls between two terms, which is exactly why the formula carries a division by .
The same reasoning works on any AP, not just the naturals. The sum is always the number of terms times the average of the first and last, which is what says. The pairing trick and the average rule are one method described from two sides — and for an odd-length list you can see the average sitting in the middle of the row.
Exam relevance
How is the arithmetic progression tested in JEE Main?
Because the AP is the first of the standard progressions, and its two formulas are assumed without restatement in every later series chapter.
This is the foundation for Class 11 Mathematics Sequences and Series and Class 10 Arithmetic Progressions, both examined in JEE Main. The formulas found here — and — are used in exactly that form, and the pairing derivation is the standard proof asked for.
**The sum of the first naturals becomes a building block.** Class 11 gives the standard summations
and the first is this page's formula. The third is its square, which is a pleasant fact worth noticing: the sum of the first cubes equals the square of the sum of the first naturals. Questions asking for the sum of a series whose th term is a polynomial in are solved by splitting into these three, and they are a recurring JEE Main type.
The arithmetic mean comes straight from the pairing idea. Class 11 defines the AM of and as — the middle term of a three-term AP — and the inequality pairs it with the geometric mean from the next part of this chapter. That inequality is a standard tool for maxima and minima problems where calculus would be slower.
Where the counting arguments lead. The handshake sum is , the number of ways to choose two things from , which Class 11 Permutations and Combinations treats directly — and the double-counting check used on this page (each handshake counted from both ends) is the standard justification for the division by in that formula.
A common exam construction. Questions often give two terms of an AP, such as the third and the seventh, and ask for a third quantity. The method here — count the steps between them, divide the gap by the number of steps to get , then work back to — handles every version of it.
What the questions look like. For board work, expect **test whether a sequence is an AP and state and , find a specified term, find which term equals a given value, derive by pairing, and a growing-pattern word problem needing a term or a total. For JEE Main, expect sums of series, problems mixing AP with GP, and the AM-GM inequality.
How board and competitive emphasis differ. A board paper rewards the written derivation** and the identified and . A competitive paper assumes both formulas and tests whether you can recognise an AP inside a word problem, or split a summation into the standard pieces.
The single trap that costs the most marks. Using instead of . For the twentieth term is , and the wrong version gives — a plausible number that is one step too far. **The defence is to test the formula at every time you write it**: it must return itself, and returns instead.
This is the foundation for Class 11 Mathematics Sequences and Series and Class 10 Arithmetic Progressions, both examined in JEE Main. The formulas found here — and — are used in exactly that form, and the pairing derivation is the standard proof asked for.
**The sum of the first naturals becomes a building block.** Class 11 gives the standard summations
and the first is this page's formula. The third is its square, which is a pleasant fact worth noticing: the sum of the first cubes equals the square of the sum of the first naturals. Questions asking for the sum of a series whose th term is a polynomial in are solved by splitting into these three, and they are a recurring JEE Main type.
The arithmetic mean comes straight from the pairing idea. Class 11 defines the AM of and as — the middle term of a three-term AP — and the inequality pairs it with the geometric mean from the next part of this chapter. That inequality is a standard tool for maxima and minima problems where calculus would be slower.
Where the counting arguments lead. The handshake sum is , the number of ways to choose two things from , which Class 11 Permutations and Combinations treats directly — and the double-counting check used on this page (each handshake counted from both ends) is the standard justification for the division by in that formula.
A common exam construction. Questions often give two terms of an AP, such as the third and the seventh, and ask for a third quantity. The method here — count the steps between them, divide the gap by the number of steps to get , then work back to — handles every version of it.
What the questions look like. For board work, expect **test whether a sequence is an AP and state and , find a specified term, find which term equals a given value, derive by pairing, and a growing-pattern word problem needing a term or a total. For JEE Main, expect sums of series, problems mixing AP with GP, and the AM-GM inequality.
How board and competitive emphasis differ. A board paper rewards the written derivation** and the identified and . A competitive paper assumes both formulas and tests whether you can recognise an AP inside a word problem, or split a summation into the standard pieces.
The single trap that costs the most marks. Using instead of . For the twentieth term is , and the wrong version gives — a plausible number that is one step too far. **The defence is to test the formula at every time you write it**: it must return itself, and returns instead.
Key takeaways
Arithmetic progressions and the sum formula: quick revision
- A sequence is an AP when consecutive differences are all equal; that value is the common difference .
- is an AP with , . has . has .
- is not an AP (differences ); nor is , which has a constant ratio instead.
- Check every difference: passes the first test and fails the second.
- From two terms: third term , seventh term gives , so and .
- **th term**: — the fourth term takes three steps.
- of is ; of is .
- Locating a value: gives ; gives .
- is not a term of , since has no whole solution.
- Sum by pairing: write the sum forwards and backwards, get columns of , then halve.
- ****: gives ; gives ; gives .
- **A sum not starting at **: , confirmed as .
- Sum of any AP: . For to the twentieth term, .
- **First odd numbers sum to **: . **First even numbers sum to **: .
- The formula always gives a whole number, since one of , is even.
- These sums are the triangular numbers: , .
- Chairs at tables: gives , so seats.
- Brick stack : rows and bricks.
- Salary rising a year: . Logs down to : .
- Handshakes among people: , confirmed as after correcting the double count.
- Savings of rising monthly for months: .
- Read whether a TERM or a TOTAL is wanted — bricks against .
Count the rows of seats in your classroom, check whether they grow by a constant amount, and if they do, predict the total without counting every chair.
- is an AP with , . has . has .
- is not an AP (differences ); nor is , which has a constant ratio instead.
- Check every difference: passes the first test and fails the second.
- From two terms: third term , seventh term gives , so and .
- **th term**: — the fourth term takes three steps.
- of is ; of is .
- Locating a value: gives ; gives .
- is not a term of , since has no whole solution.
- Sum by pairing: write the sum forwards and backwards, get columns of , then halve.
- ****: gives ; gives ; gives .
- **A sum not starting at **: , confirmed as .
- Sum of any AP: . For to the twentieth term, .
- **First odd numbers sum to **: . **First even numbers sum to **: .
- The formula always gives a whole number, since one of , is even.
- These sums are the triangular numbers: , .
- Chairs at tables: gives , so seats.
- Brick stack : rows and bricks.
- Salary rising a year: . Logs down to : .
- Handshakes among people: , confirmed as after correcting the double count.
- Savings of rising monthly for months: .
- Read whether a TERM or a TOTAL is wanted — bricks against .
Count the rows of seats in your classroom, check whether they grow by a constant amount, and if they do, predict the total without counting every chair.