Pair the Numbers From Both Ends and a Long Addition Becomes One Multiplication
Add the first n terms of an arithmetic progression with either sum formula, work backwards from a given sum to find n, recover any term from the sums, and solve savings, salary, seat, log-stack and prize-money problems.
How do you add up a hundred numbers without adding them one by one?
Suppose you have to add . Done term by term that is ninety-nine additions and a near-certainty of error.
But pair the numbers from the two ends inward and something remarkable happens:
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Every pair gives the same total, because as one number climbs by the other falls by . There are such pairs, so the whole sum is
One multiplication in place of ninety-nine additions. And the pairing trick works for any arithmetic progression, for exactly the same reason: the terms rise by a constant from one end and fall by the same from the other, so the two movements cancel in every pair.
Turning that idea into a formula is what this part of the chapter does, and it gives you two versions of the same result — one for when you know the first term and the common difference, one for when you know the first and last terms.
From there the questions run in four directions.
- Given the AP, find the sum of a stated number of terms
- **Given the sum, find , or — which usually produces a quadratic equation
- Given the sums, recover a single term**, using
- Model a real situation — money saved month by month, seats in widening rows, logs stacked in a heap, prizes decreasing in value
This page covers the second part of the CBSE Class 10 Maths chapter on arithmetic progressions: the sum of the first terms, working backwards from a given sum, and applied problems.
But pair the numbers from the two ends inward and something remarkable happens:
-
-
-
Every pair gives the same total, because as one number climbs by the other falls by . There are such pairs, so the whole sum is
One multiplication in place of ninety-nine additions. And the pairing trick works for any arithmetic progression, for exactly the same reason: the terms rise by a constant from one end and fall by the same from the other, so the two movements cancel in every pair.
Turning that idea into a formula is what this part of the chapter does, and it gives you two versions of the same result — one for when you know the first term and the common difference, one for when you know the first and last terms.
From there the questions run in four directions.
- Given the AP, find the sum of a stated number of terms
- **Given the sum, find , or — which usually produces a quadratic equation
- Given the sums, recover a single term**, using
- Model a real situation — money saved month by month, seats in widening rows, logs stacked in a heap, prizes decreasing in value
This page covers the second part of the CBSE Class 10 Maths chapter on arithmetic progressions: the sum of the first terms, working backwards from a given sum, and applied problems.
Formula
What are the two formulas for the sum of an arithmetic progression?
**Use the first if you know and ; use the second if you know the first and last terms.** They give the same answer, and having both is what makes this topic quick.
where is the last term of the terms being added, that is .
Where the second one comes from. It is the pairing argument written down: the average of the first and last terms is , and every pair from the two ends has that same average, so the total is the number of terms times that average. The sum of an AP is the number of terms multiplied by the average of the first and last — worth saying in words, because it makes the formula impossible to misremember.
**And the first formula is the second one with replaced by , so there is really only one result here, written two ways.
Worked example 1 — both formulas on the same problem.** Find the sum of the first terms of the AP with and .
Using the first formula:
Using the second, which needs the last term first:
The same answer by two routes — a genuine check, and it costs one extra line.
Worked example 2 — a decreasing AP. Find the sum of the first terms of .
Here and :
Check with the other formula: , so
Correct. A negative sum is perfectly normal once the terms have gone below zero.
Worked example 3 — summing a given list. Find .
This is an AP with , and . First find how many terms there are:
Then
**Notice that .** The sum of the first odd numbers is always , which is one of the neatest results in the chapter and a useful thing to recognise.
The step students skip. When a question gives you the last term rather than the number of terms, **you must find first.** A sum formula needs to know how many terms it is adding, and there is no way around that short calculation.
where is the last term of the terms being added, that is .
Where the second one comes from. It is the pairing argument written down: the average of the first and last terms is , and every pair from the two ends has that same average, so the total is the number of terms times that average. The sum of an AP is the number of terms multiplied by the average of the first and last — worth saying in words, because it makes the formula impossible to misremember.
**And the first formula is the second one with replaced by , so there is really only one result here, written two ways.
Worked example 1 — both formulas on the same problem.** Find the sum of the first terms of the AP with and .
Using the first formula:
Using the second, which needs the last term first:
The same answer by two routes — a genuine check, and it costs one extra line.
Worked example 2 — a decreasing AP. Find the sum of the first terms of .
Here and :
Check with the other formula: , so
Correct. A negative sum is perfectly normal once the terms have gone below zero.
Worked example 3 — summing a given list. Find .
This is an AP with , and . First find how many terms there are:
Then
**Notice that .** The sum of the first odd numbers is always , which is one of the neatest results in the chapter and a useful thing to recognise.
The step students skip. When a question gives you the last term rather than the number of terms, **you must find first.** A sum formula needs to know how many terms it is adding, and there is no way around that short calculation.
How do you work backwards from a given sum to find n, a or d?
**Substitute everything you know into a sum formula and solve for what is left. When is the unknown the equation is usually quadratic, and one root must often be rejected.
Worked example 1 — finding the number of terms.** How many terms of the AP must be taken to give a sum of ?
Here , and :
By the quadratic formula,
giving or .
**A number of terms cannot be negative or fractional, so .
Check**: . Correct.
Worked example 2 — finding the first term. The sum of the first terms of an AP is and the common difference is . Find the first term.
Check by listing the seven terms: . Adding them gives . Correct.
Worked example 3 — recovering a term from the sums. The sum of the first terms of a sequence is . Find its nth term, its first term and its tenth term.
The key relation is
Why it works. is the total of the first terms and the total of the first , so their difference is the single term that was added last — the nth term. So
So , giving , , and .
Two checks, and both should be done. First, must equal : . Correct. Second, , and the sum formula gives
Correct. And directly, . All three agree.
**Notice what the shape of told you.** is a quadratic in , and its nth term came out linear in — which is the signature of an AP. A quadratic sum always means an AP, and the common difference is twice the coefficient of , here , which matches .
**The danger with .** It is valid only for , because there is no term to subtract when . Always confirm the first term separately as , and check that the general formula agrees with it — as it did above.
Worked example 1 — finding the number of terms.** How many terms of the AP must be taken to give a sum of ?
Here , and :
By the quadratic formula,
giving or .
**A number of terms cannot be negative or fractional, so .
Check**: . Correct.
Worked example 2 — finding the first term. The sum of the first terms of an AP is and the common difference is . Find the first term.
Check by listing the seven terms: . Adding them gives . Correct.
Worked example 3 — recovering a term from the sums. The sum of the first terms of a sequence is . Find its nth term, its first term and its tenth term.
The key relation is
Why it works. is the total of the first terms and the total of the first , so their difference is the single term that was added last — the nth term. So
So , giving , , and .
Two checks, and both should be done. First, must equal : . Correct. Second, , and the sum formula gives
Correct. And directly, . All three agree.
**Notice what the shape of told you.** is a quadratic in , and its nth term came out linear in — which is the signature of an AP. A quadratic sum always means an AP, and the common difference is twice the coefficient of , here , which matches .
**The danger with .** It is valid only for , because there is no term to subtract when . Always confirm the first term separately as , and check that the general formula agrees with it — as it did above.
How do savings, seats, logs and prize problems turn into sums?
Whenever a quantity changes by the same amount at each step, the running total is the sum of an AP. Name , and from the story, then use a sum formula.
Worked example 1 — savings. A man saves ₹ in the first month and increases his saving by ₹ every month. How much does he save in a year?
Here , and :
**He saves ₹ in the year.
Check with the other formula**: the twelfth month's saving is , so . Correct.
Worked example 2 — seats in rows. The first row of a hall has seats and each row after it has more seats than the row before. How many seats are there in rows?
**There are seats.
Check**: the fifteenth row has seats, so . Correct.
Worked example 3 — prize money, where the first term is unknown. A sum of ₹ is to be given as seven cash prizes, each prize being ₹ less than the one before it. Find the value of each prize.
This is worked example 2 of the previous section: , so the prizes are **₹, ₹, ₹, ₹, ₹, ₹ and ₹.
Notice how the question was phrased. The total was given and the first term had to be found, which is the reverse of the savings problem. Read which quantity is missing before choosing a formula.
Worked example 4 — a stack of logs, where a root must be rejected.** Two hundred logs are stacked so that there are logs in the bottom row, in the row above, in the next, and so on. In how many rows are the logs placed, and how many logs are in the top row?
Here , and :
giving or .
Both are positive whole numbers, so neither can be discarded on sight. Test each against the situation:
- **If **, the top row would have logs. Impossible
- **If **, the top row has logs. Perfectly sensible
**So there are rows and the top row holds logs.
Check**: . Correct.
This is the most instructive rejection in the whole chapter. Both roots satisfied the equation, and the arithmetic gave no hint which to keep. The test was a quantity the question never mentioned — the number of logs in the top row — and finding it required going back to the meaning of the AP. **Whenever a sum question gives two admissible values of , compute the last term for each.
Worked example 5 — an annual increment.** A person's salary in the first year of a job is ₹ and it rises by ₹ each year. Find the total earned over years.
**The total is ₹.
Check**: the tenth year's salary is , so . Correct.
Worked example 1 — savings. A man saves ₹ in the first month and increases his saving by ₹ every month. How much does he save in a year?
Here , and :
**He saves ₹ in the year.
Check with the other formula**: the twelfth month's saving is , so . Correct.
Worked example 2 — seats in rows. The first row of a hall has seats and each row after it has more seats than the row before. How many seats are there in rows?
**There are seats.
Check**: the fifteenth row has seats, so . Correct.
Worked example 3 — prize money, where the first term is unknown. A sum of ₹ is to be given as seven cash prizes, each prize being ₹ less than the one before it. Find the value of each prize.
This is worked example 2 of the previous section: , so the prizes are **₹, ₹, ₹, ₹, ₹, ₹ and ₹.
Notice how the question was phrased. The total was given and the first term had to be found, which is the reverse of the savings problem. Read which quantity is missing before choosing a formula.
Worked example 4 — a stack of logs, where a root must be rejected.** Two hundred logs are stacked so that there are logs in the bottom row, in the row above, in the next, and so on. In how many rows are the logs placed, and how many logs are in the top row?
Here , and :
giving or .
Both are positive whole numbers, so neither can be discarded on sight. Test each against the situation:
- **If **, the top row would have logs. Impossible
- **If **, the top row has logs. Perfectly sensible
**So there are rows and the top row holds logs.
Check**: . Correct.
This is the most instructive rejection in the whole chapter. Both roots satisfied the equation, and the arithmetic gave no hint which to keep. The test was a quantity the question never mentioned — the number of logs in the top row — and finding it required going back to the meaning of the AP. **Whenever a sum question gives two admissible values of , compute the last term for each.
Worked example 5 — an annual increment.** A person's salary in the first year of a job is ₹ and it rises by ₹ each year. Find the total earned over years.
**The total is ₹.
Check**: the tenth year's salary is , so . Correct.
How do you sum the natural numbers and the multiples between two limits?
**The natural numbers form an AP with and , so the sum formula collapses to something you can quote.**
Deriving it takes one line from with and , which is exactly the pairing argument of the opening section.
Worked example 1. Find the sum of the first natural numbers.
Check by pairing: pairs each summing to , so . Correct.
For multiples between two limits there are three steps, and the first is the one that goes wrong.
- Find the first multiple inside the range and the last multiple inside the range — not the limits themselves
- **Find ** from
- Apply a sum formula
Worked example 2. Find the sum of all multiples of lying between and .
The first multiple of above is and the last below is , so , , :
**The sum is .
Check by factoring out the **: the multiples are . Correct, and that shortcut is worth knowing — the sum of the first multiples of is times the sum of the first natural numbers.
Worked example 3. Find the sum of all multiples of lying between and .
The first multiple of above is ; the last below is . So , , :
**The sum is .
Check with the other formula**: . Correct.
The trap in both of those is the word "between". It excludes the limits, so with multiples of between and you start at and not at , and you stop at and not at . **If the question says "from to " or "inclusive", check whether either limit is itself a multiple** — here is not a multiple of , so the answer would be unchanged, but and must still be identified deliberately rather than assumed.
One more result that follows from the same idea. The sum of the first odd numbers is , as the sum showed. **And the sum of the first even numbers is **, since . Both are one-line consequences of the natural-number sum, and both are quick to check on a small case: the first three even numbers add to , and .
Deriving it takes one line from with and , which is exactly the pairing argument of the opening section.
Worked example 1. Find the sum of the first natural numbers.
Check by pairing: pairs each summing to , so . Correct.
For multiples between two limits there are three steps, and the first is the one that goes wrong.
- Find the first multiple inside the range and the last multiple inside the range — not the limits themselves
- **Find ** from
- Apply a sum formula
Worked example 2. Find the sum of all multiples of lying between and .
The first multiple of above is and the last below is , so , , :
**The sum is .
Check by factoring out the **: the multiples are . Correct, and that shortcut is worth knowing — the sum of the first multiples of is times the sum of the first natural numbers.
Worked example 3. Find the sum of all multiples of lying between and .
The first multiple of above is ; the last below is . So , , :
**The sum is .
Check with the other formula**: . Correct.
The trap in both of those is the word "between". It excludes the limits, so with multiples of between and you start at and not at , and you stop at and not at . **If the question says "from to " or "inclusive", check whether either limit is itself a multiple** — here is not a multiple of , so the answer would be unchanged, but and must still be identified deliberately rather than assumed.
One more result that follows from the same idea. The sum of the first odd numbers is , as the sum showed. **And the sum of the first even numbers is **, since . Both are one-line consequences of the natural-number sum, and both are quick to check on a small case: the first three even numbers add to , and .
Exam tip
Which habits protect a sum-of-an-AP answer?
**Write down , and before touching a formula, and finish by checking with the other formula. The two sum formulas are an in-built verification and almost nobody uses them that way.
- State , and explicitly**, with the sign of
- **If the last term is given rather than , find first — a sum formula cannot run without it
- Use when you know both ends**, and otherwise
- Check with the other formula. If they disagree, the error is in the arithmetic, not the method
- **When solving for , expect a quadratic and reject any root that is negative or fractional
- If both roots of are positive integers, compute the last term for each — the log-stack problem is decided that way, not by the equation
- Confirm ** whenever you use , since that relation needs
- For multiples "between" two limits, identify the first and last multiple inside the range, never the limits themselves
- **Quote for the naturals**, and remember the odd numbers sum to
- Give units and answer in words — ₹, seats, rows
The misconception to name. The sum of an AP is not the number of terms times the middle term in general. It is the number of terms times the average of the first and last, and those coincide only when is odd. For the sum is , the average of the ends is , and — but there is no single middle term to use.
A second trap. Using where is wanted, or the reverse. **A question asking how much he saves in the tenth month wants ; one asking how much he has saved by the tenth month wants .** Reading the preposition is genuinely part of the mathematics here, and the two answers are nowhere near each other.
- State , and explicitly**, with the sign of
- **If the last term is given rather than , find first — a sum formula cannot run without it
- Use when you know both ends**, and otherwise
- Check with the other formula. If they disagree, the error is in the arithmetic, not the method
- **When solving for , expect a quadratic and reject any root that is negative or fractional
- If both roots of are positive integers, compute the last term for each — the log-stack problem is decided that way, not by the equation
- Confirm ** whenever you use , since that relation needs
- For multiples "between" two limits, identify the first and last multiple inside the range, never the limits themselves
- **Quote for the naturals**, and remember the odd numbers sum to
- Give units and answer in words — ₹, seats, rows
The misconception to name. The sum of an AP is not the number of terms times the middle term in general. It is the number of terms times the average of the first and last, and those coincide only when is odd. For the sum is , the average of the ends is , and — but there is no single middle term to use.
A second trap. Using where is wanted, or the reverse. **A question asking how much he saves in the tenth month wants ; one asking how much he has saved by the tenth month wants .** Reading the preposition is genuinely part of the mathematics here, and the two answers are nowhere near each other.
Did you know
Why does the sum of an AP always come out as a quadratic in n?
Expand the sum formula and look at what you are left with:
**A quadratic in with no constant term. That shape is not an accident, and it explains several things at once.
Why there is no constant term.** Putting must give , because adding no terms gives nothing. A quadratic that passes through the origin has no constant, and any sum formula you derive that ends up with one is wrong.
Why the nth term comes out linear. The nth term is the difference , and the difference of consecutive values of a quadratic is always linear. So the sums of an AP are quadratic and the terms are linear — one degree apart, always.
Which gives a fast way to read an AP off its sum. If , then:
- **The common difference is twice the coefficient of
- The first term is the sum of the two coefficients**, since
**Test it on .** The coefficient of is , so ; the coefficients add to , so . **And the AP really is , exactly as the earlier working found. Two answers read straight off the expression, with no differencing at all.
The same pattern continues upward, and it is worth knowing exists.** If a sequence's terms are quadratic in , its sums are cubic; the differences of the terms are then linear, and the differences of those are constant. Taking differences lowers the degree by one each time, which is why the constant-difference test identifies an AP in the first place: an AP's terms are degree one, so one round of differencing flattens them to a constant.
And that is why the pairing trick worked at the very start. Pairing from the ends is really the observation that does not depend on — the rise in one term exactly offsets the fall in the other, because both move linearly. For a sequence of squares the trick fails, and the failure is informative: but , so no constant pair-sum exists and the sum of squares needs a different formula altogether.
**A quadratic in with no constant term. That shape is not an accident, and it explains several things at once.
Why there is no constant term.** Putting must give , because adding no terms gives nothing. A quadratic that passes through the origin has no constant, and any sum formula you derive that ends up with one is wrong.
Why the nth term comes out linear. The nth term is the difference , and the difference of consecutive values of a quadratic is always linear. So the sums of an AP are quadratic and the terms are linear — one degree apart, always.
Which gives a fast way to read an AP off its sum. If , then:
- **The common difference is twice the coefficient of
- The first term is the sum of the two coefficients**, since
**Test it on .** The coefficient of is , so ; the coefficients add to , so . **And the AP really is , exactly as the earlier working found. Two answers read straight off the expression, with no differencing at all.
The same pattern continues upward, and it is worth knowing exists.** If a sequence's terms are quadratic in , its sums are cubic; the differences of the terms are then linear, and the differences of those are constant. Taking differences lowers the degree by one each time, which is why the constant-difference test identifies an AP in the first place: an AP's terms are degree one, so one round of differencing flattens them to a constant.
And that is why the pairing trick worked at the very start. Pairing from the ends is really the observation that does not depend on — the rise in one term exactly offsets the fall in the other, because both move linearly. For a sequence of squares the trick fails, and the failure is informative: but , so no constant pair-sum exists and the sum of squares needs a different formula altogether.
Exam relevance
How is the sum of a progression used in JEE?
This is foundation work for Class 11 Sequences and Series, one of the most consistently examined algebra chapters in JEE Main and a regular source of JEE Advanced problems.
Where the sum formula leads. Class 11 sets it beside the sum of a geometric progression and the sums of , and , and uses all of them together to sum series whose general term is a polynomial. The method — find the general term, then sum it using known results — begins with the AP case you learn here.
**Where leads.** It is the standard way of recovering a general term from a sum, and JEE Main sets it directly: a sum is given as an expression in and the nth term is required. **The caution that the relation holds only for is exactly what such a question tests**, because a candidate who forgets to check can produce a formula that fails at the first term.
Where the quadratic-sum observation leads. Reading and off the coefficients of a quadratic sum is a genuine competitive shortcut, and the general principle — that summing raises the degree by one and differencing lowers it — underlies the method of differences used for harder series.
Where the natural-number sums lead. is used in Binomial Theorem and Permutations and Combinations, and the fact that the first odd numbers sum to appears in counting arguments. Both are quoted, not derived, at that level.
Where the word problems lead. Class 12 Application of Derivatives and the arithmetic-reasoning sections reuse the same modelling: a quantity changing by a fixed amount per step, with a total to be optimised or matched. The log-stack problem, where two roots satisfy the equation and only one fits the situation, is the shape of a great many competitive questions.
Question types to expect. At this level: find a sum, find from a sum, recover a term from , and a full word problem. In competitive papers: summing series built from AP and GP terms, finding a general term from a given sum, means inserted between numbers, and problems mixing a progression with a quadratic condition.
The single trap that costs marks. Confusing with . **"How much in the tenth month" is ; "how much by the tenth month" is — and at JEE the same confusion appears as summing a general term that was itself already a sum.
A second trap.** Accepting both roots of without testing them. **The log stack gives and , and implies a row with logs — the equation cannot detect that, only the meaning can. Competitive papers rely on this kind of hidden constraint, and the guard is always to compute the last term.
Board versus competitive emphasis.** The CBSE paper marks the values of , and , the formula, the substitution and the interpretation; a competitive paper marks the number or the general term. The transferable habit is checking every sum with the second formula — it takes one line, it uses information you already have, and it turns a plausible answer into a certain one.
Where the sum formula leads. Class 11 sets it beside the sum of a geometric progression and the sums of , and , and uses all of them together to sum series whose general term is a polynomial. The method — find the general term, then sum it using known results — begins with the AP case you learn here.
**Where leads.** It is the standard way of recovering a general term from a sum, and JEE Main sets it directly: a sum is given as an expression in and the nth term is required. **The caution that the relation holds only for is exactly what such a question tests**, because a candidate who forgets to check can produce a formula that fails at the first term.
Where the quadratic-sum observation leads. Reading and off the coefficients of a quadratic sum is a genuine competitive shortcut, and the general principle — that summing raises the degree by one and differencing lowers it — underlies the method of differences used for harder series.
Where the natural-number sums lead. is used in Binomial Theorem and Permutations and Combinations, and the fact that the first odd numbers sum to appears in counting arguments. Both are quoted, not derived, at that level.
Where the word problems lead. Class 12 Application of Derivatives and the arithmetic-reasoning sections reuse the same modelling: a quantity changing by a fixed amount per step, with a total to be optimised or matched. The log-stack problem, where two roots satisfy the equation and only one fits the situation, is the shape of a great many competitive questions.
Question types to expect. At this level: find a sum, find from a sum, recover a term from , and a full word problem. In competitive papers: summing series built from AP and GP terms, finding a general term from a given sum, means inserted between numbers, and problems mixing a progression with a quadratic condition.
The single trap that costs marks. Confusing with . **"How much in the tenth month" is ; "how much by the tenth month" is — and at JEE the same confusion appears as summing a general term that was itself already a sum.
A second trap.** Accepting both roots of without testing them. **The log stack gives and , and implies a row with logs — the equation cannot detect that, only the meaning can. Competitive papers rely on this kind of hidden constraint, and the guard is always to compute the last term.
Board versus competitive emphasis.** The CBSE paper marks the values of , and , the formula, the substitution and the interpretation; a competitive paper marks the number or the general term. The transferable habit is checking every sum with the second formula — it takes one line, it uses information you already have, and it turns a plausible answer into a certain one.
Key takeaways
What must you be able to do from this part?
Two formulas for one result, and a rejection test the equation cannot do for you.
- **** when you know and ; ** when you know both ends
- The sum is the number of terms times the average of the first and last — which is the pairing argument in words
- Check every sum with the other formula.** With , , both give
- **Find first** whenever the last term is given instead — has terms and sums to
- **Solving for gives a quadratic.** For summing to , the equation gives
- ****, valid for only, so always confirm
- **** gives , with and
- A quadratic sum always means an AP: is twice the coefficient of , and is the sum of the two coefficients
- ****, so the first naturals sum to
- **The first odd numbers sum to ** and the first even numbers to
- For multiples between limits, take the first and last multiple inside the range — multiples of between and sum to , and multiples of between and sum to
- **Savings of ₹ rising by ₹ a month total ₹ in a year; rows starting at seats and rising by hold seats; ₹ in seven prizes falling by ₹ starts at ₹
- Two hundred logs from a bottom row of decreasing by one** fill rows with in the top row — is rejected because it would need logs
- Read the preposition: "in the tenth month" is , "by the tenth month" is
The best self-test is the double check. Take any AP you like, sum its first twelve terms with one formula and then with the other, and see whether the two lines land on the same number without repeating a single multiplication.
- **** when you know and ; ** when you know both ends
- The sum is the number of terms times the average of the first and last — which is the pairing argument in words
- Check every sum with the other formula.** With , , both give
- **Find first** whenever the last term is given instead — has terms and sums to
- **Solving for gives a quadratic.** For summing to , the equation gives
- ****, valid for only, so always confirm
- **** gives , with and
- A quadratic sum always means an AP: is twice the coefficient of , and is the sum of the two coefficients
- ****, so the first naturals sum to
- **The first odd numbers sum to ** and the first even numbers to
- For multiples between limits, take the first and last multiple inside the range — multiples of between and sum to , and multiples of between and sum to
- **Savings of ₹ rising by ₹ a month total ₹ in a year; rows starting at seats and rising by hold seats; ₹ in seven prizes falling by ₹ starts at ₹
- Two hundred logs from a bottom row of decreasing by one** fill rows with in the top row — is rejected because it would need logs
- Read the preposition: "in the tenth month" is , "by the tenth month" is
The best self-test is the double check. Take any AP you like, sum its first twelve terms with one formula and then with the other, and see whether the two lines land on the same number without repeating a single multiplication.