Weighted partial sums and changing averages

All lessons · Armenia study route · Start adaptive practice

Your goal: Count each entry’s contribution to running totals, track changes in their average, and prove comparisons and optimal orders. You should be able to explain why the position of a number matters even when the total stays fixed.

Preparation context: This original lesson is calibrated to Grade 9 school-stage mathematics preparation in Armenia. The released 2025–26 Grade 9 school paper includes reasoning about averages of partial sums. Its objective questions are a useful target for efficient calculation; the written proofs below are explicitly labelled bridges towards longer mathematical arguments. This is a topic lesson, not a full paper or a prediction of an upcoming question.

Study plan: Use two sittings of about 30–40 minutes for the diagnostic, theory and four examples. Then use separate sessions for the written problems. Try a nearby task before moving on. Revisit any solution-assisted problem later without opening the explanation.

Prerequisites: Exact arithmetic and fractions, collecting like terms, simple linear equations, and sequence notation. No inequality theorem or advanced summation formula is assumed.

Three short prerequisite checks

Attempt these before opening the solutions. If the arithmetic mean is unclear, repair Check 1 first. If repeated terms are difficult to collect, use Check 2 as the starting point for the contribution table.

Check 1. What is being averaged?

Three numbers have values 4, 10 and 7. Find their arithmetic mean. If three numbers instead have mean 8, what is their total?

Complete solution

The arithmetic mean is the total divided by the number of entries. Here $4 + 10 + 7 = 21$, so the mean is $21/3 = 7$. Reversing the operation, a mean of 8 for three entries means a total of $3 \times 8 = 24$. Always identify how many quantities are being averaged.

Check 2. Count copies before simplifying

Collect like terms in $p + (p + q) + (p + q + r)$.

Complete solution

The first bracket-free term supplies one copy of $p$; the next group supplies another, and the last supplies a third. There are two copies of $q$ and one of $r$. Thus the expression is $3p + 2q + r$. This counts repeated contributions; it does not assume any numerical values or signs for the letters.

Check 3. Undo a weighted total

Solve $4x + 18 = 42$, and check your answer.

Complete solution

Subtract 18 from each side to obtain $4x = 24$. Divide by 4 to get $x = 6$. Substitution gives $4 \times 6 + 18 = 42$, so the value satisfies the original equation. In this lesson, the coefficient 4 will often mean that an entry has been counted four times.

One entry can contribute several times

Write an ordered list of real numbers as $a_{1}, a_{2}, \ldots , a_{n}$, where $n$ is a positive integer. A partial sum, or running total, is the sum from the first entry up to a chosen position. Thus $S_{1} = a_{1}$, $S_{2} = a_{1} + a_{2}$, and in general $S_{k} = a_{1} + \cdots + a_{k}$. The final total is $T = S_{n}$.

We will study $A$, the arithmetic mean of the n running totals. It is different from the mean of the original entries. First add the running totals, then divide by their number:

$A=\frac{S_1+\cdots+S_n}{n}$

For four entries, lay out the contributions. A dash means that the entry has not yet been included, not that the entry itself is zero.

Read down the columns. The first entry occurs four times, the second three times, the third twice and the fourth once. Adding by columns gives the same total as adding by rows:

$S_{1} + S_{2} + S_{3} + S_{4} = 4a_{1} + 3a_{2} + 2a_{3} + a_{4}$

Why the general coefficient is one more than you may expect. Entry $a_{k}$ appears in $S_k,S_{k+1},\ldots,S_n$. The included positions run from $k$ through $n$, so there are $n - k + 1$ of them. For example, the positions 3 through 6 are 3, 4, 5, 6: four positions, not three.

$nA=na_1+(n-1)a_2+\cdots+2a_{n-1}+a_n$

This identity is valid even when some entries are zero or negative. We are counting copies, not assuming that the running totals increase. For $n = 1$, it simply says $A = a_{1}$.

Do not change the denominator. The coefficients count appearances in the numerator. We still average exactly $n$ running totals, so the denominator is $n$. Dividing by the sum of the coefficients would instead give a weighted mean of the original entries, which is a different quantity.

Example 1. Two routes to the same average

The entries, in order, are 3, 1, 4 and 2. Find the average of their four running totals.

Start directly. The running totals are 3, 4, 8 and 10, so their average is $\frac{3+4+8+10}{4}=\frac{25}{4}$.

Now count contributions. The first entry appears four times and the last once. Therefore the numerator is $4 \times 3 + 3 \times 1 + 2 \times 4 + 2 = 25$. Dividing by 4 gives the same answer, $\frac{25}{4}$. The ordinary mean of the entries is $\frac{10}{4}$, which answers a different question.

Your turn 1. Make both calculations agree

The entries are 2, 5, 1 and 4. Find the average of the running totals by two methods.

Hint 1

Write all four running totals first.

Hint 2

Compare their sum with the weighted numerator $4 \times 2 + 3 \times 5 + 2 \times 1 + 4$.

Complete solution

The running totals are 2, 7, 8 and 12, with sum 29. Thus the average is $\frac{29}{4}$. The contribution calculation gives $4 \times 2 + 3 \times 5 + 2 \times 1 + 4 = 29$, again followed by division by 4.

Track a change without rebuilding every total

Suppose only entry $a_{k}$ changes by an amount $\delta$. Every running total before position $k$ stays the same. Each of the remaining $n - k + 1$ totals changes by $\delta$. Consequently:

$\text{change in }A=\frac{(n-k+1)\delta}{n}$

A negative change works by the same reasoning. As a boundary check, changing the first entry by 1 changes every running total by 1, so $A$ changes by 1. Changing the last entry by 1 changes only one running total, so $A$ changes by $\frac{1}{n}$.

Example 2. Locate the changed entry

A list has six real entries. Exactly one entry is increased by 12. The average of its six running totals rises by 8. Which position was changed?

Recover the change in the numerator. The average rose by 8 across six quantities, so the sum of running totals rose by $6 \times 8 = 48$. Each affected running total rose by 12. Thus exactly $48/12 = 4$ running totals were affected.

Translate a count into a position. In a list of six entries, an entry at position $k$ occurs in $7 - k$ running totals. Therefore $7 - k = 4$, giving $k = 3$. Check: the third entry occurs in totals 3, 4, 5 and 6. Their combined increase is 48, and $48/6 = 8$ is the required average increase. The ordinary mean of the entries rises by only $12/6 = 2$.

Your turn 2. A smaller influence identifies a later position

A list has eight real entries. Increasing one entry by 20 raises the average of its eight running totals by 5. Find the position.

Hint 1

First undo the average by multiplying by 8.

Hint 2

Divide the change in the sum of running totals by 20 to count the affected totals.

Complete solution

The sum of the running totals increases by $8 \times 5 = 40$. Since each affected total rises by 20, exactly two totals are affected. Position 7 occurs in totals 7 and 8, so the changed entry is the seventh. Equivalently, $9 - k = 2$ gives $k = 7$.

Bridge: moving an amount instead of adding it. Suppose an amount $\delta$ is transferred from a later position $j$ to an earlier position $i$, with $i < j$. The earlier entry increases by $\delta$ and the later entry decreases by $\delta$, so the final total is unchanged. Running totals from position $i$ through $j - 1$ rise by $\delta$; from position $j$ onwards, the two changes cancel. Exactly $j - i$ totals rise. Hence:

$\text{change in }A=\frac{(j-i)\delta}{n}$

This is often easier than comparing two long weighted expressions. If a problem restricts entries to nonnegative numbers, check that the transfer does not make the donating entry negative.

Reverse the list and pair complementary weights

Let $B$ be the average of the running totals after reversing the list. The final total $T$ stays the same. An entry at original position $k$ moves to position $n + 1 - k$. Its weight in the original numerator is $n + 1 - k$; in the reversed numerator its weight is $k$. These add to $n + 1$, regardless of the position.

$nA + nB = (n + 1)T$

$A+B=\frac{(n+1)T}{n}$

For four entries, the weight pairs are $4 + 1$, $3 + 2$, $2 + 3$ and $1 + 4$. Each entry is counted five times across the two numerators. This proves the identity for arbitrary real entries; a symmetric-looking list is not required.

Example 3. Find an average without knowing the entries

Five real entries have total 18. In their given order, the average of the running totals is 12. Find the corresponding average after reversing the order.

The forward and reversed averages satisfy $A+B=\frac{6\times18}{5}$, because each entry has combined weight 6 across the two numerators. Therefore $B=\frac{108}{5}-12=\frac{48}{5}$.

We did not need to reconstruct the five entries. To check that the conditions are possible, use 10, 0, 1, 0, 7. Its total is 18, and its running totals 10, 10, 11, 11, 18 sum to 60, giving average 12. In reverse order the running totals are 7, 7, 8, 8, 18, which sum to 48 and average $\frac{48}{5}$. This example checks attainability; the identity proves the answer for every list satisfying the given conditions.

Your turn 3. Use both pieces of information

Four real entries have total 20. Their forward running-total average is 14. Find the reversed running-total average.

Hint 1

The forward and reversed averages add to a quantity determined by the final total.

Hint 2

For four entries, the combined weight of every entry is 5.

Complete solution

Each original entry has combined weight 5 in the two numerators, so $A+B=\frac{5\times20}{4}=25$. With $A = 14$, the reversed average is $B = 11$.

Which order makes the average largest?

For a fixed collection of entries, the weights are largest at the beginning. To justify placing larger values earlier, compare only two affected positions. Suppose $i < j$, with values $x$ and $y$ there. Interchanging them changes the numerator by:

$(j - i)(y - x)$

To see why, the difference between the two position weights is $j - i$. Moving $y$ earlier gains that many copies of $y$, while moving $x$ later loses that many copies of $x$. If $x < y$, this change is positive. Thus any smaller entry placed before a larger one can be swapped to increase the average.

From a useful swap to a complete proof. Take the largest available entry and put it first by a swap, if necessary. Such a swap never decreases the numerator. Then put a largest remaining entry second, and continue. In at most one placement per position, the list is in nonincreasing order. Therefore that order attains the maximum. Placing a smallest remaining entry in each position gives a nondecreasing list and the minimum. This argument works with negative entries and with repeated values. Exchanging equal entries has no effect.

This is an elementary exchange argument. We have proved the part needed here rather than assuming the general rearrangement inequality.

Example 4. Optimise, including a negative entry

Arrange −2, 1, 4 and 7 to make the average of the four running totals as large as possible and as small as possible. Give both averages and justify the orders.

Maximum. The weights are 4, 3, 2, 1. The exchange argument puts the largest entry first, giving the order 7, 4, 1, −2. Its numerator is $4 \times 7 + 3 \times 4 + 2 \times 1 - 2 = 40$, so the maximum average is $40/4 = 10$.

Minimum. Put the entries in increasing order: −2, 1, 4, 7. Its numerator is $4 \times (-2) + 3 \times 1 + 2 \times 4 + 7 = 10$, so the minimum average is $\frac{10}{4}=\frac{5}{2}$.

Why no other order does better. If a smaller value comes before a larger one, swapping that pair strictly increases the numerator, so a maximum must have descending order. Reversing the argument gives the ascending minimum. The values here are distinct, so these value orders are unique. As a check, their total is 10 and $10+\frac{5}{2}=\frac{5\times10}{4}$, as the reverse identity requires.

Your turn 4. What changes when values repeat?

Arrange 1, 3, 3 and 5 to maximise and minimise the running-total average. Explain what exchanging the two 3s does.

Hint 1

Match larger entries to larger coefficients for a maximum.

Hint 2

The change from swapping equal entries contains the factor $3 - 3 = 0$.

Complete solution

The maximum value order is 5, 3, 3, 1, giving numerator $4 \times 5 + 3 \times 3 + 2 \times 3 + 1 = 36$ and average 9. The minimum value order is 1, 3, 3, 5, giving numerator $4 \times 1 + 3 \times 3 + 2 \times 3 + 5 = 24$ and average 6. The two 3s are equal, so interchanging them changes no value and no running total. The ordered lists of values are fixed, although artificially labelling the equal copies would distinguish their labels.

A bridge from calculation to comparison

Sometimes the entries are already increasing and cannot be rearranged. Compare the forward and reversed averages by pairing opposite positions. With four entries, subtracting the two weighted numerators gives:

$4(A - B) = 3(a_{1} - a_{4}) + (a_{2} - a_{3})$

If $a_{1} \le a_{2} \le a_{3} \le a_{4}$, both differences on the right are nonpositive. Hence $A \le B$. Their sum is already known, so $A$ is at most half that sum. This pairing extends to any length: the first and last entries have opposite difference-coefficients, then the second and second-last do, and so on. For an odd length the middle entry has coefficient zero. Written Problem 7 asks you to complete this proof and its equality case.

Choose the first step from the information

• Entries are given: form running totals or count each entry’s copies.
• One position changes: count affected totals; there are $n - k + 1$.
• An amount moves between positions: only the totals between those positions change.
• The order is reversed: pair complementary weights and use the final total.
• An arrangement must be best: prove what a swap does, then justify that an ordered arrangement is reached.
• A fixed total is distributed among positive integers: reserve the mandatory 1 in each position before allocating what remains.

Errors to catch before practising

• Averaging the entries when the question asks for the average of running totals.
• Giving the last entry weight zero. It occurs in the final running total, so its weight is 1.
• Dividing the weighted numerator by the sum of its weights instead of by the number of running totals.
• Assuming a fixed final total determines the running-total average; the positions also matter.
• Claiming “largest first” without showing that a swap improves the objective.
• Using a nonnegative or positive-integer restriction only at the end. It can change the attainable minimum.
• Treating equality of the forward and reversed averages as proof that the list is a palindrome. Problem 6 supplies a counterexample.

Nine original written problems

Problems 1–2 establish the representation, Problems 3–6 require transfer and complete reasoning, and Problems 7–9 are proof bridges. These are teaching tiers, not official paper sections or an official marking scheme. Try each before opening Hint 1; use Hint 2 only if the first does not unlock a next step.

Foundation · 1. Negative entries do not break the count

The five entries are 4, −1, 3, 0 and 2. Find the average of their five running totals, and verify it by contribution counting.

Hint 1

Compute the next running total from the preceding one, including the negative change.

Hint 2

Use weights 5, 4, 3, 2, 1 for a second calculation.

Complete solution

The running totals are 4, 3, 6, 6 and 8, whose sum is 27. The average is $\frac{27}{5}$. By contributions, the same numerator is $5 \times 4 + 4 \times (-1) + 3 \times 3 + 2 \times 0 + 2 = 27$. The negative entry lowers every running total from position 2 onwards; its weight is still 4.

Foundation · 2. Recover two missing entries

Four real entries are $a, 2, b, 1$, in that order. Their final total is 12 and the average of their running totals is $\frac{33}{4}$. Find $a$ and $b$.

Hint 1

The final total and the running-total sum give different equations.

Hint 2

Multiply the given average by 4, and use weights 4, 3, 2, 1.

Complete solution

The final total gives $a + b = 9$. The sum of the four running totals is 33, so contribution counting gives $4a + 6 + 2b + 1 = 33$, or $4a + 2b = 26$. Twice the total equation is $2a + 2b = 18$. Subtraction gives $2a = 8$, hence $a = 4$ and $b = 5$.

The resulting list 4, 2, 5, 1 has running totals 4, 6, 11, 12. They sum to 33 and finish at 12. The two linear equations force the values, so no other pair satisfies both conditions.

Core · 3. A transfer preserves one average but changes another

Eight real entries have total 30. Increase the second entry by 3 and decrease the seventh entry by 3. Find the changes in the ordinary mean of the entries and in the mean of their running totals.

Hint 1

List the running totals in which the increase has occurred but the decrease has not.

Hint 2

The affected positions are 2 through 6, inclusive.

Complete solution

The two entry changes add to zero, so the total remains 30 and the ordinary mean remains $\frac{30}{8}=\frac{15}{4}$. Its change is zero.

The running total at position 1 is unchanged. Totals at positions 2, 3, 4, 5 and 6 rise by 3. At positions 7 and 8, the decrease of 3 cancels the earlier increase. Thus exactly five totals rise by 3, and the running-total average rises by $\frac{5\times3}{8}=\frac{15}{8}$. There is no nonnegativity restriction here, so the transfer is permitted for any original real entries.

Core · 4. Two directions and an attainable example

Five nonnegative integer entries have total 15. Their forward running-total average exceeds their reversed running-total average by 6. Find both averages and give one possible list.

Hint 1

Use the final total to find the sum of the two averages.

Hint 2

Solve their sum and difference; then build an example with most of the total early.

Complete solution

Let the forward and reversed averages be $A$ and $B$. The reverse identity gives $A+B=\frac{6\times15}{5}=18$. We also know $A - B = 6$. Adding the equations gives $2A = 24$, so $A = 12$ and $B = 6$.

It remains to show that the integer and sign conditions can really be met. Take the list 9, 3, 0, 0, 3. Its total is 15, and its running totals 9, 12, 12, 12, 15 sum to 60, giving average 12. In reverse order the running totals are 3, 3, 3, 6, 15; they sum to 30 and average 6. Every entry is a nonnegative integer. Thus the computed averages are attainable.

Core · 5. Find every pair of changed positions

In a list of six real entries, one entry is increased by 6 and a different entry is decreased by 6. The average of the six running totals increases by 4. Find all possible ordered pairs (increased position, decreased position).

Hint 1

Use the two position weights even before deciding which position comes first.

Hint 2

The change in the average simplifies to the decreased position minus the increased position.

Complete solution

Call the increased position $i$ and the decreased position $j$. Their combined effect on the weighted numerator is $6(7 - i) - 6(7 - j) = 6(j - i)$. After dividing by 6, the increase in the average is $j - i$. Therefore $j - i = 4$.

Both positions lie between 1 and 6, so $j = i + 4 \le 6$ forces $i \le 2$. Thus the pairs are $(1,5)$ and $(2,6)$. Each has distance 4 and therefore raises four running totals by 6, producing average increase 4. The position bounds exclude every other pair; in particular, the increased position must be earlier.

Core · 6. Equal averages do not force a symmetric list

Arrange the four distinct numbers 1, 2, 5 and 8 so that the average of the four running totals is 10. Find every possible order.

Hint 1

The total is fixed. What must the reversed average be if the forward average is 10?

Hint 2

For four entries, equality of the two averages gives $3(a - d) + (b - c) = 0$. Bound the endpoint difference.

Complete solution

Write the order as $a,b,c,d$. The total is 16, so the reverse identity gives $A+B=\frac{5\times16}{4}=20$. Thus $A = 10$ is equivalent to $A = B$. Subtracting the two numerators gives $3(a - d) + (b - c) = 0$. In particular, the absolute difference between the middle entries must be three times that between the endpoints.

The difference between any two available values is at most 7. Therefore the endpoint difference is at most $\frac{7}{3}$. Among 1, 2, 5 and 8, the only pair with positive difference at most this bound is 1 and 2. The middle entries must then be 5 and 8. If $a = 1$ and $d = 2$, the equation requires $b - c = 3$, so the order is 1, 8, 5, 2. Reversing it gives the other order 2, 5, 8, 1.

The first order has running totals 1, 9, 14, 16, summing to 40; the second has 2, 7, 15, 16, also summing to 40. Each averages 10. The endpoint argument proves completeness. Neither list is a palindrome, because its first and last entries differ.

Self-check: Show equivalence to the difference equation, restrict the endpoints, check the two orders, and explain why none was missed.

Proof bridge · 7. Increasing entries give an upper bound

Let $n \ge 2$ be an integer and let $a_{1} \le a_{2} \le \cdots \le a_{n}$ be real numbers with total $T$. Prove that the average $A$ of their running totals satisfies $A\le\frac{(n+1)T}{2n}$. Determine exactly when equality holds.

Hint 1

Compare the list with its reverse and pair opposite positions in the difference.

Hint 2

Each paired contribution is a positive coefficient times an earlier entry minus a later entry. Use the reverse sum identity after proving $A \le B$.

Complete solution

Let $B$ be the reversed running-total average. In $nA - nB$, the entry at position $k$ has coefficient $(n + 1 - k) - k = n + 1 - 2k$. Its opposite entry at position $n + 1 - k$ has coefficient $k - (n + 1 - k) = -(n + 1 - 2k)$. Group these two opposite coefficients: for each $k<\frac{n+1}{2}$, the pair contributes $(n+1-2k)(a_k-a_{n+1-k})$.

The coefficient is positive and, because the entries are nondecreasing, the difference is nonpositive. Every pair therefore contributes at most zero. If there is a middle entry, its coefficient is zero. Adding all contributions gives $n(A - B) \le 0$, hence $A \le B$.

Now $2A\le A+B=\frac{(n+1)T}{n}$, which gives the required bound after division by 2. No sign assumption on $T$ was needed.

Equality. All pair contributions are nonpositive. If their sum is zero, the outermost pair, whose coefficient is $n - 1 > 0$, must be zero. Therefore $a_{1} = a_{n}$. Since every intermediate entry lies between them, all entries are equal. Conversely, if every entry equals a real number $c$, the forward list equals its reverse, so $A = B$ and the bound is attained. Thus equality holds exactly when all entries are equal.

Self-check: Explain every paired sign, handle a possible middle entry, use the known sum, and prove both directions of the equality condition.

Proof bridge · 8. A fixed total with positive integer entries

Six positive integers have total 21. Their values and order may both be chosen. Find the smallest and largest possible average of their six running totals. Prove both bounds and give lists attaining them.

Hint 1

Separate the compulsory positive part from the units you can move.

Hint 2

After placing a 1 in every position, 15 units remain. Each contributes at least 1 and at most 6 to the weighted numerator.

Complete solution

Reserve one unit in each position, because every entry must be positive. The reserved list 1, 1, 1, 1, 1, 1 uses total 6 and contributes $6 + 5 + 4 + 3 + 2 + 1 = 21$ to the weighted numerator. There are $21 - 6 = 15$ units left to distribute, each with a weight between 1 and 6.

Minimum. Those 15 units contribute at least 15, so the numerator is at least $21 + 15 = 36$. This gives average at least 6. Put all extra units in the last position: 1, 1, 1, 1, 1, 16. Its running totals are 1, 2, 3, 4, 5, 21, summing to 36, so it attains 6.

Maximum. The extra units contribute at most $6 \times 15 = 90$, so the numerator is at most $21 + 90 = 111$, giving average at most $\frac{111}{6}=\frac{37}{2}$. The list 16, 1, 1, 1, 1, 1 attains this bound: its running totals are 16, 17, 18, 19, 20, 21, summing to 111. All entries in both constructions are positive integers with total 21.

Self-check: Reserve the mandatory units, bound every remaining contribution, divide by six, and verify constructions attaining both bounds.

Proof bridge · 9. Delete the first entry without changing the average

A list has five positive integer entries. After its first entry is deleted, the average of the four new running totals is the same as the average of the five original running totals. Find the smallest possible common average and prove your answer.

Hint 1

Name the four new running totals. How does each relate to an original running total?

Hint 2

The old sum equals five copies of the deleted entry plus the new sum. Use the same average with two different denominators.

Complete solution

Let the first entry be $a$ and the common average be $C$. Denote the four running totals after deletion by $R_{1}, R_{2}, R_{3}, R_{4}$. The five original running totals are $a, a + R_{1}, a + R_{2}, a + R_{3}, a + R_{4}$. Their sum is therefore $5a + R_{1} + R_{2} + R_{3} + R_{4}$.

The new running-total sum is $4C$, while the old one is $5C$. Consequently $5C = 5a + 4C$, so $C = 5a$. Since $a$ is a positive integer, $a \ge 1$ and therefore $C \ge 5$.

To attain this lower bound, take 1, 1, 1, 1, 11. Its running totals are 1, 2, 3, 4, 15, which sum to 25 and average 5. After deleting the first 1, the running totals are 1, 2, 3, 14, which sum to 20 and also average 5. All five entries are positive integers. Thus the smallest possible common average is exactly 5.

Self-check: Relate the old and new totals explicitly, use positivity for a lower bound, and provide a valid list attaining it.

Write a complete argument

For a proof or optimisation problem, use this editorial self-check. It is not an official marking scheme and is not automatic grading of a written proof.

1. Define the quantities. State the number of entries, their domain, and which average is required.
2. Justify the weights. Identify the running totals containing a chosen entry or affected by a change.
3. Explain the decisive step. A swap needs a sign calculation; a paired comparison needs the sign of every contribution.
4. Establish completeness. For “all orders”, exclude unlisted orders. For a minimum or maximum, prove a bound and then attain it.
5. Check equality and constraints. Verify positivity, integrality, total, and any equality case in the original question.

Adaptive checks with a return route

The objective questions start with contribution counts and progress to reverse-order and optimisation reasoning. The practice responds to your answers with hints, a complete explanation and a link to the relevant teaching section. The engine does not grade arbitrary written proofs.

A completed session means six responses, including any requested solutions. Completion is not a claim of independent mastery. After assisted work, return to a fresh attempt later. Lessons, hints and solutions are free; the existing optional browser-local progress and privacy choices continue to apply.

Interactive practice loads here. You can also use the complete question set below.

Explain before you return to a paper

Without looking back, explain why the last entry has weight 1 and why the first has weight equal to the list length. Then describe which running totals change when an amount is moved from a later position to an earlier one. Finally, choose a written problem you solved with help and attempt it again without its hints.

Return to Armenia’s study and official papers · Review sequences and sums · Explore the general rearrangement inequality · Choose another lesson

Preparation sources and original material

The released 2025–26 Grade 9 school-stage paper on Armenia’s official Olympiad portal provides the topic and preparation-level context. Its Question 20 involves averages of partial sums. Read the Armenia country guide and current organiser notices for the actual competition route and current instructions. The longer proof tasks here are teaching extensions, not a claim that the school paper requires this proof format.

All explanations, tables, examples and practice questions on this page are original IMOolympiad.com teaching material. Standard mathematical identities are proved in the lesson. No official question or textbook exercise is reproduced, translated or adapted by changing its numbers. Official papers are linked for reading at the source. This independent educational site is not the organiser and claims no endorsement or affiliation. Preparation context checked 5 November 2026.