NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

Decide in one glance: does it add, or does it scale?

• The test: ask whether the same rupee or unit amount gets added every step, or whether the quantity gets scaled. A flat add is an AP; a percentage or fraction of what is left is not.
• Flat adds: in (i) the rule is plus  8 per km and in (iii) plus  50 per metre. The phrases "each additional" and "rises by" signal a constant add, so both are APs.
• Shrinking add: in (ii) the pump takes a fraction of the air that remains, so the amount removed shrinks each time and the step is never constant, so it fails the test.
• Growing percent: in (iv) compound interest grows the money by a fixed percentage, not a fixed rupee amount, so the yearly increase keeps rising and it is not an AP.
• Why it matters: examiners mix a compound-interest trap into such lists to see if students confuse constant addition with constant ratio. Spotting add versus scale earns the full reasoning marks.

(i) and (iii) are APs (constant add); (ii) and (iv) are not (constant ratio).

Build a tiny table, then read it off.

• Set the slots: for each part write the four slots a, a+d, a+2d, a+3d first, then plug in the given values rather than adding in your head. This is how toppers avoid silent slips.
• Zero is allowed: in (ii) the difference is zero, so the AP is the constant list -2,-2,-2,-2. Many students wrongly assume the difference must be non-zero.
• Keep the form: in (iv) and (v) hold the fractions and decimals exactly as given and never round, so the run -1.25, -1.50, -1.75, -2.00 stays clean.
• Why it matters: generating terms from the first value and the difference is the base skill for every later question, so fixing the signs and the zero case here prevents a chain of errors in harder problems.

Add d to the previous term three times; the four terms come straight out.

Confirm the difference with a third term.

• Double-check rule: the fast way to be sure of the difference is to also compute the next gap and see it matches the first.
• Part (i): the first gap is -2 and the next gap -1-1 is also -2. They agree, so the difference is safely -2.
• Part (iii): keeping a common denominator over three lets you read the difference off instantly as four-thirds.
• Part (iv): with decimals just subtract carefully, since both 1.7-0.6 and 2.8-1.7 give 1.1.
• Why it matters: reading the first term and difference correctly from a printed list is the first line of almost every AP solution, so a quick second check removes careless slips before they spread.

a is the first number listed; d = a₂ - a₁, confirmed by a₃ - a₂.

Group the lookalike traps by type before computing.

• Three families: the fifteen lists hide three classic traps, which are powers that are really geometric, surds that must be simplified first, and squares or decimals that only pretend to be regular.
• Geometric traps: parts (i), (ix) and (xi) multiply by a fixed factor, so their gaps grow instead of staying constant, and you can reject all three at once without much work.
• Surd traps: part (xii) cleans up to a multiple of root two and is a genuine AP, while part (xiii) collapses to terms whose gaps are unequal, so simplify every root before you judge the pattern.
• Square traps: the odd squares in (xiv) give rising gaps and fail, whereas (xv) keeps a constant gap of twenty-four once you read its last entry as the plain number seventy-three rather than a square.
• Why it matters: examiners pack this question with lookalikes to test whether you actually compute differences instead of guessing from the shape, so sorting them into powers, surds and squares handles all fifteen quickly and correctly.

Compute consecutive differences (after simplifying surds): nine are APs, six are not.

Treat each row as a mini equation.

• The routine: the fastest, error-free route is to substitute the known values first and rearrange second, so the unknown ends up alone on one side.
• Row (ii): the unknown is the difference, so move the constant across to get 9d = 18 and then divide to reach two.
• Row (iii): the unknown is the first term, so add fifty-one to both sides and the answer is forty-six.
• Row (iv): the unknown is the term number, so isolate the bracket as nine and then add one back, a step students often forget.
• Why it matters: filling such a table tests whether you can rearrange the same formula for any missing quantity, which is exactly the skill needed for the longer word problems later in the chapter.

Substitute the three known values into a_n = a + (n-1)d and solve for the blank.

Eliminate options by sign before you compute.

• The idea: in an MCQ you can often cut choices before computing fully, because the sign of the difference already tells you whether the target term is large positive or large negative.
• Part (i): the difference is negative, so the thirtieth term sits well below the start and only the two negative options survive, and computing the value confirms option (C).
• Part (ii): read the difference carefully because the second term is a fraction; a positive difference pushes the eleventh term well above the start, so both negative options are out.
• Finish (ii): the arithmetic then lands exactly on option (B).
• Why it matters: these one-mark questions reward speed, so reading the sign first lets you reject half the options before any arithmetic and the formula only has to confirm the survivor.

(i) -77, option (C); (ii) 22, option (B).

Count the gaps, not the boxes.

• Core rule: the number of difference-steps between two known terms is always the later position minus the earlier position, and once you get that right the common difference falls out immediately.
• Part (iv): here the ends are term one and term six, so there are five steps, the difference is two, and filling forward gives the run -4,-2,0,2,4,6.
• Part (v): the known terms are positions two and six, so four steps give a difference of minus fifteen; step back once for the first term and forward for the rest.
• Parts (ii) and (iii): use the known term plus the right number of steps to the last term to pin the difference, then fill the boxes one step at a time.
• Why it matters: miscounting the gap by one is the classic slip, so anchoring on "position difference equals number of steps" makes every box reliable and earns method marks even before the final numbers.

Find d from two known terms (steps = position gap), then add d to fill each box.

Read it as how many steps of five.

• Find the rise: the total climb from the first term three up to the target seventy-eight is seventy-five.
• Count the steps: each step adds five, so seventy-five divided by five is fifteen steps from the first term.
• Add the start: the term sits one place beyond the first because the first term needs zero steps, so the position is sixteen.
• Why it matters: seeing the bracket as total rise over step size turns the formula into a plain division, which is faster under exam pressure and makes the extra one for the first term obvious.

The 16th term equals 78.

Convert the mixed number before you subtract.

• Clear the trap: in part (ii) the mixed number is the snag, so turn it into an improper fraction first, which makes the difference come out exactly as minus five-halves.
• Part (i): the total span from seven to two hundred five is one hundred ninety-eight; dividing by the step six gives thirty-three steps, so there are thirty-four terms.
• Part (ii): the span from eighteen down to minus forty-seven is minus sixty-five; dividing by the exact difference gives twenty-six steps, so there are twenty-seven terms.
• Whole-number check: always confirm the count is a whole number, because if it is not then the listed value is not actually a term.
• Why it matters: counting terms is the base for the sum formula and for "is this a term" questions, so an exact difference from the mixed number keeps the whole later chain correct.

(i) 34 terms; (ii) 27 terms.

Set up the equation, then judge the term number.

• Whole idea: the test rests on whether the term number is a positive integer, so do the algebra fully and inspect the final fraction.
• Rearrange: write the value as the general term and move the first term across to get minus one hundred sixty-one equal to the step times the bracket.
• Divide and inspect: dividing leaves the bracket as one hundred sixty-one over three, which is clearly not a whole number.
• State the reason: conclude the value is not a term and write "the term number is not a whole number" to earn the justification mark.
• Why it matters: boards ask "is this a term" to check that you know an AP only takes integer positions, and showing the non-integer count is the exact proof the examiner is looking for.

Not a term; n works out to a fraction, not a positive integer.

Use the gap between the two given terms.

• Read the gap: from the eleventh to the sixteenth term is five steps, and the value rises by thirty-five, so the difference is seven in a single line, d = a₁₆ - a₁₁16 - 11 = 355 = 7.
• Jump forward: from the eleventh to the thirty-first term is twenty more steps of seven each, so the thirty-first term is one hundred seventy-eight.
• Skip the first term: this route never needs the actual first term, which is handy whenever it is not asked.
• Why it matters: the difference of terms over difference of positions gives the common difference instantly and is a clean way to jump between any two terms without solving for the first, saving time in long papers.

a₃₁ = 178.

Gap method for the difference, then one jump.

• Find the step: from term three to term fifty is forty-seven steps over a rise of ninety-four, so the difference is two, d = a₅₀ - a₃50 - 3 = 9447 = 2.
• Jump to target: from term three to term twenty-nine is twenty-six steps, so the twenty-ninth term is sixty-four.
• No first term: there is no need to compute the first term when only the twenty-ninth is wanted.
• Why it matters: jumping straight from a known term to the target using the step count avoids an extra equation and is less error-prone in a fifty-term AP where the numbers are large.

a₂₉ = 64.

Track the sign change to spot the zero.

• Set the scene: the third term is positive and the ninth is negative, so the terms must cross zero somewhere between positions three and nine.
• Find the step: the gap method gives a difference of minus two, d = a₉ - a₃9 - 3 = -126 = -2, so each term drops by two.
• Walk to zero: from the third term, two more drops of two reach zero, which carries you across positions three, four and five.
• Read the answer: the fifth term is therefore zero, matching the full algebra.
• Why it matters: watching where positive turns to negative gives a quick sanity check and builds intuition for AP behaviour, which helps in sum problems where some terms cancel.

The 5th term is zero.

Skip the first term completely.

• Key fact: the first term never appears in the difference of two terms, so there is no need to even introduce it here.
• Write the gap: straight from the formula the difference of the seventeenth and tenth terms is seven times the common difference.
• Match and solve: set this equal to the given excess of seven and divide, which gives a common difference of one.
• Why it matters: many "exceeds by" problems fall in two lines once you know the difference of terms is just the position gap times the common difference, which saves writing a full pair of equations.

d = 1.

Use the step-count shortcut.

• Spot the step: each term here jumps by twelve, so gaining a fixed amount over a known term is just counting how many twelve-steps fit.
• Count the steps: one hundred thirty-two divided by twelve is eleven steps.
• Place the term: the wanted term is eleven positions after the fifty-fourth, which is the sixty-fifth, matching the full equation in one division.
• Why it matters: reading "more by an amount" as a number of steps turns a long substitution into a quick division, a real time saver when the term numbers are large.

The 65th term.

No numbers needed, just cancellation.

• Key insight: equal common differences make the variable parts of the two terms identical, so they subtract to zero and only the gap between the first terms survives.
• It is constant: that surviving difference does not depend on the term number, so the hundredth and thousandth differences must be equal.
• Read the answer: since the hundredth difference is given as one hundred, the thousandth difference is one hundred too.
• Why it matters: this question checks whether you see that the common difference controls the spacing while the first term controls the starting offset, and the offset is exactly what survives at any term number.

100, unchanged, because the difference equals a - a' for every term.

Count by division at both ends.

• The trick: a neat alternative is to count all multiples of seven up to nine hundred ninety-nine and subtract those up to ninety-nine.
• Upper count: the floor of nine hundred ninety-nine over seven is one hundred forty-two multiples.
• Lower count: the floor of ninety-nine over seven removes the fourteen one-digit and two-digit multiples.
• Subtract: the three-digit multiples are one hundred forty-two minus fourteen, which is one hundred twenty-eight and matches the AP count.
• Why it matters: this floor-division method is a quick cross-check on the AP answer and is handy whenever you do not want to write out the full progression.

128 such numbers.

Span over step, then add one.

• The rule: the number of terms equals the span between the first and last terms divided by the step size, plus one for the starting term.
• Find the span: the distance from the first multiple twelve to the last multiple two hundred forty-eight is two hundred thirty-six.
• Count the steps: dividing that span by the step of four gives fifty-nine steps.
• Add the start: adding one for the first term gives sixty multiples in all.
• Why it matters: the span over step plus one rule is the heart of every counting-in-a-range problem, and internalising it makes these one-mark questions almost instant.

60 multiples.

Read it as one AP catching the other.

• Set the race: the first AP starts sixty ahead but grows slower, so the second is steadily catching up.
• Initial lead: the first term of the leading AP is sixty above the first term of the other.
• Closing rate: the faster AP gains five on the slower one every single term.
• Catch-up point: the lead of sixty is wiped out after twelve steps, which lands them equal at the thirteenth term.
• Why it matters: reading the problem as a faster AP catching a slower one gives the same answer through simple division and builds a feel for how common differences compete.

n = 13.

Get the difference from the excess first.

• Pick the lone unknown: the difference of the seventh and fifth terms involves only the common difference, so solve for it before touching the first term.
• Solve for the step: the excess of twelve equals twice the difference, so the difference is six immediately.
• Use the other fact: the remaining condition that the third term is sixteen then gives a first term of four.
• Present the AP: list a few terms as four, ten, sixteen, twenty-two and so on.
• Why it matters: solving for the unknown that appears alone keeps the algebra simple and is a good habit for any two-condition AP problem in the exam.

4, 10, 16, 22, …

Or count the terms, then convert the position.

• Two valid routes: some students prefer to reverse the AP, while others find the total number of terms first and then convert the asked position to one counted from the front, and both are correct.
• Count the terms: solving the last-term equation gives a total of fifty-one terms in the progression.
• Convert the position: the twentieth term from the last is the same as the thirty-second term from the front, since you flip the position about the total count.
• Compute the value: the thirty-second term from the front works out to one hundred fifty-eight, exactly the same answer as the reverse method.
• Why it matters: having two independent routes that agree lets you pick whichever feels safer, with the reverse method quicker and the count-then-convert method a reliable check that guards against a careless slip.

158.

Subtract the paired sums to isolate the step.

• Why it works: the two sum-equations differ only in their difference coefficients, so a single subtraction wipes out the first-term part and leaves the step alone.
• Find the step: the second equation minus the first gives the common difference as five.
• Back-substitute: putting that step into the first equation gives the first term as minus thirteen.
• List the terms: adding the step twice produces the first three terms, minus thirteen, minus eight and minus three.
• Why it matters: sum-of-terms conditions reduce to neat linear equations once each term is expanded, and subtracting to remove the first-term part is the standard clean route examiners expect.

-13, -8, -3.

Count the fixed increments of two hundred.

• Frame it: the salary climbs by a fixed raise each year, so the whole question is just how many raises take the starting pay up to the target.
• Total rise: the climb from five thousand to seven thousand is two thousand.
• Count raises: dividing that rise by the yearly raise of two hundred gives ten increments.
• Place the year: ten increments after the start year of nineteen ninety-five lands in two thousand five.
• Why it matters: translating a salary problem into how many fixed raises makes the arithmetic a one-line division and keeps the year count straight, which is exactly how such applied questions are marked.

Year 2005.

Clear the decimals before dividing.

• The idea: working with the decimal step is cleaner once you scale both numbers up to whole numbers.
• Rise in savings: the increase from five to twenty point seven five is fifteen point seven five.
• Count the steps: dividing by the weekly step gives nine once you multiply top and bottom by a hundred to clear the decimals.
• Add the start: adding the first week gives a total of ten weeks.
• Why it matters: many savings and instalment problems use decimal increments, and scaling to integers before dividing keeps the work neat and reliably avoids arithmetic mistakes in the exam.

n = 10.

Match the denominators in the fraction AP.

• Spot the hard part: the first three sums are routine, and only the fraction AP in part (iv) needs care, so choose a single common denominator of sixty up front and the rest becomes plain arithmetic.
• Find the step: subtracting the first two terms over the common denominator gives a difference of one over sixty.
• Add inside the bracket: rewrite the leading fraction over sixty so it adds cleanly to the rest, which simplifies to three over ten.
• Finish the sum: multiplying by the half-count factor gives the final value of thirty-three over twenty.
• Why it matters: sum problems with fractions are really a test of denominator handling, and converting everything to a single denominator before adding is the safest route to the exact answer the board expects.

(i) 245; (ii) -180; (iii) 5505; (iv) 3320.

Two-stage routine: count, then average.

• Same shape: every "up to a last term" sum follows the same two stages, so handle all three parts with one routine.
• Stage one, count: use the last-term equation to find the number of terms, which gives twenty-three, thirteen and seventy-six.
• Stage two, average: average the two ends and scale by the count, S_n = n × a+l2; for part (iii) the average term is -5+(-230)2, and times seventy-six gives minus eight thousand nine hundred thirty.
• Read the form: this half-count times sum-of-ends is simply the number of terms times the average term.
• Why it matters: reading the sum as count times average term demystifies the formula and gives a built-in sanity check, since the sum should sit near the count times the middle value.

(i) 104612; (ii) 286; (iii) -8930.

Reject any negative or fractional term count.

• Watch the quadratics: parts (vi) and (viii) turn into quadratics in the term count, and only a positive whole number is a valid count.
• Part (vi): the quadratic factors cleanly, so reject the negative root, keep five, and the fifth term is thirty-four.
• Part (viii): again the quadratic factors, so reject the negative root, keep seven, and the first term is minus eight.
• The rest: everywhere else just match the known quantities to the right formula and solve linearly, taking extra care with the fractions in parts (ii) and (v).
• Why it matters: the sum formula can produce a quadratic in the count, and knowing the count must be a positive integer lets you discard the extra root instantly, a step examiners specifically check.

Solve the matching formula for the unknowns; discard any non-positive or non-integer n.

Tidy the quadratic before you solve it.

• The crux: the whole question turns on reducing the quadratic carefully and then recognising that its discriminant is a perfect square.
• Form the equation: the sum formula gives the tidy quadratic 4n² + 5n - 636 = 0.
• Check the discriminant: the discriminant works out to ten thousand two hundred one, which is exactly the square of one hundred one.
• Take the valid root: this gives a count of twelve once the negative root is rejected, n = -5 + 1018 = 12.
• Why it matters: a clean perfect-square discriminant signals you are on track, and spotting it confirms the count without a calculator, exactly the kind of number boards design.

n = 12 terms.

Read the sum as count times average term.

• The shortcut: the sum equals the count times the average of the two ends, so the count appears at sight once you know that average.
• Average term: the average of the first and last terms is twenty-five.
• Count the terms: dividing the total of four hundred by that average of twenty-five gives sixteen terms.
• Find the step: the last-term equation then gives the common difference as eight-thirds.
• Why it matters: treating the sum as count times average term makes the count obvious and gives a quick check, since sixteen terms averaging twenty-five indeed add to four hundred.

n = 16, d = 83.

Span over step for the count, then average.

• Find the count: the total rise from seventeen to three hundred fifty is three hundred thirty-three, and dividing by the step of nine gives thirty-seven steps, so there are thirty-eight terms.
• Average the ends: the average of the first and last terms is one hundred eighty-three point five.
• Scale to the sum: multiplying that average by the thirty-eight terms gives six thousand nine hundred seventy-three, matching the formula.
• Why it matters: splitting the work into counting the terms and then averaging them gives two simple computations and a natural cross-check on the final sum.

38 terms; sum = 6973.

Two clean steps: find the start, then average.

• Find the first term: the twenty-second-term equation collapses to a first term of two.
• Average the ends: the average of the first and last terms is seventy-five point five.
• Scale to the sum: multiplying that average by the twenty-two terms gives one thousand six hundred sixty-one.
• Why it matters: once the last term is identified as the twenty-second term, the sum needs only the first term, which one equation supplies, keeping a seemingly long problem to two lines.

S₂₂ = 1661.

Build the first term and step, then plug in.

• The pivot: the whole problem rests on getting the first term right from the given second term.
• Find the step: the two consecutive terms give a common difference of four.
• Step back: moving one place back from the second term gives a first term of ten.
• Plug into the sum: the sum formula for fifty-one terms then gives five thousand six hundred ten.
• Why it matters: a single misread, using the second term as the first, would throw off the entire sum, so stepping back to the true first term is the safeguard examiners reward.

S₅₁ = 5610.

Notice the perfect squares in the data.

• Early hint: the two given sums are the squares of seven and seventeen, which already hints that the sum of the first count of terms may equal the count squared.
• Reduce the conditions: dividing each condition by its count gives the tidy pair that yields a step of two and a first term of one.
• Get the closed form: substituting these back collapses the sum formula to the count squared.
• Check it: the closed form reproduces both given sums exactly, matching the perfect squares.
• Why it matters: finding a clean closed form lets you write any partial sum instantly and is a satisfying confirmation that the given data were consistent.

S_n = n².

The coefficient of the term number is the step.

• Key shortcut: when the term formula is linear in the term number, the coefficient of that number is automatically the common difference.
• Part (i): the term coefficient is four, so the difference is four and the first term is seven.
• Part (ii): the term coefficient is minus five, so the difference is minus five and the first term is four.
• The sums: the fifteen-term sums then come out as five hundred twenty-five and minus four hundred sixty-five.
• Why it matters: recognising that any linear term formula is always an AP with that coefficient as the difference lets you read the step off instantly and focus your working on the sum.

(i) AP with d = 4, S₁₅ = 525; (ii) AP with d = -5, S₁₅ = -465.

Difference the sum to get the term formula.

• Clean route: the tidy way to the general term is the difference between successive sums, which collapses the quadratic neatly.
• Shift the sum: compute the sum to one fewer term, which expands to a simple quadratic in the term number.
• Subtract: the difference of the two sums cancels the square term and leaves the linear term formula five minus twice the count.
• Check it: that formula reproduces the first, second and tenth terms exactly, all matching the direct values.
• Why it matters: once you have the term formula, every individual term and the difference of minus two follow at once, and deriving the term formula from the sum is a core skill the board tests through this exact question type.

a_n = 5 - 2n, giving a₁ = 3, a₂ = 1, a₃ = -1, a₁₀ = -15, and S₂ = 4.

Pull the common factor out of the sum.

• The idea: the multiples of six are just six times the plain counting numbers, so factor the six out and add the counting numbers.
• Sum the counts: the first forty counting numbers add to eight hundred twenty by the well-known formula.
• Multiply back: six times eight hundred twenty is four thousand nine hundred twenty, agreeing with the direct AP formula.
• Why it matters: factoring out the common multiplier turns the problem into the familiar sum of the first forty natural numbers, a fast and reliable cross-check.

4920.

Use the same factor-out idea here.

• The idea: the first fifteen multiples of eight are just eight times the first fifteen counting numbers, so pull the eight out.
• Sum the counts: the first fifteen counting numbers add to one hundred twenty.
• Multiply back: eight times one hundred twenty is nine hundred sixty.
• Why it matters: recognising the first few multiples of a number as that number times the triangular sum is a shortcut that works for any divisor and saves plugging into the long AP formula.

960.

Use the odd-number square law.

• The law: the sum of the first count of odd numbers equals that count squared, which short-circuits the whole problem.
• Count them: the odd numbers from one to forty-nine number twenty-five in all.
• Apply it: the sum of the first twenty-five odd numbers is therefore twenty-five squared, six hundred twenty-five.
• Cross-check: this matches the full AP-sum computation exactly.
• Why it matters: the identity is worth remembering, since it answers this whole question in one line and reinforces the link between APs and perfect squares.

625.

Average daily penalty times the number of days.

• Last-day penalty: the thirtieth day's penalty is the first day plus twenty-nine raises of fifty, which is one thousand six hundred fifty.
• Average it: the average of the first and last day penalties is nine hundred twenty-five.
• Scale to total: multiplying that average by the thirty days gives twenty-seven thousand seven hundred fifty, matching the sum formula.
• Why it matters: computing the last term and averaging is a sturdy way to cross-check a penalty or instalment total, making sure you have not confused the final-day amount with the running total.

 27750.

Anchor on the middle prize.

• Find the centre: for an odd number of terms the average prize is the middle one, and the total divided by seven makes that middle prize one hundred rupees.
• Build outward: step up by twenty to the higher prizes and down by twenty to the lower ones around that centre to get the seven values.
• Sanity check: the seven prizes are symmetric about one hundred rupees and add back to the given total of seven hundred.
• Why it matters: for an odd-length AP the middle term equals the average, so spotting that the fourth prize is one hundred rupees lets you build the list outward without solving for the first term first.

 160, 140, 120, 100, 80, 60, 40.

Sum the classes first, then triple.

• The plan: it is cleaner to add the per-section counts first and multiply by the three sections only at the end.
• One section: a single section across the twelve classes plants the sum of one to twelve, which is seventy-eight trees.
• All sections: with three sections per class the total is three times seventy-eight, which is two hundred thirty-four.
• Cross-check: this matches the direct AP sum of three, six and so on up to thirty-six.
• Why it matters: pulling the section count outside the sum reduces the problem to adding the first twelve natural numbers, a quicker and less error-prone path.

234 trees.

Sum the radii, then multiply by pi.

• Factor pi out: every semicircle length is pi times its radius, so pull the pi outside all thirteen lengths and add only the radii.
• Add the radii: the thirteen radii form their own AP and sum to forty-five point five centimetres.
• Multiply back: the total length is pi times forty-five point five, using the given value of twenty-two over seven.
• Clean cancellation: the numerator becomes one thousand and one, which divides by seven to a clean one hundred forty-three centimetres.
• Why it matters: taking the common pi outside turns a geometry problem into a plain AP sum of radii, and the clean cancellation confirms the chosen value of pi was meant to give a whole number.

143 cm.

Reality-check both roots of the quadratic.

• Two candidates: the quadratic for the number of rows gives two values, sixteen and twenty-five, but only one can be physically real.
• Reject the bad root: for twenty-five rows the top row would hold minus four logs, which is meaningless.
• Keep the good root: for sixteen rows the top row holds five logs, a sensible positive count.
• Why it matters: word problems often yield two algebraic roots, but physical limits such as no negative logs decide the real answer, and checking the top-row count is the deciding step examiners look for.

16 rows; 5 logs in the top row.

Factor the doubling out of the sum.

• The idea: the total run is twice the sum of the one-way distances, so add the one-way distances first and double at the end.
• One-way AP: the one-way distances form a simple AP starting at five with a step of three over ten terms.
• One-way sum: that AP sums to one hundred eighty-five metres.
• Double it: the total run is twice one hundred eighty-five, which is three hundred seventy metres.
• Why it matters: pulling the doubling outside lets you sum the simpler one-way distances first, keeping the numbers small and the doubling explicit so the return trip is never missed.

370 m.

Find where the term crosses zero.

• Crossing point: set the term formula equal to zero and the crossing falls near a term number of thirty-one point two five, where the sign flips.
• Straddle it: the thirty-first term is just positive while the thirty-second is just negative.
• Read the answer: so the first negative term is the thirty-second.
• Why it matters: treating "first negative term" as a crossing point of the term formula makes the rounding direction obvious and guards against the off-by-one error that catches many students.

The 32nd term, equal to -3.

Centre the two terms on their average.

• The setup: since the two terms add to six their average is three, so write them as three minus twice the step and three plus twice the step, as they sit two steps either side of the middle.
• Auto-sum: written this way the two terms automatically add to six, so the sum condition is already satisfied.
• Use the product: the product becomes a difference of squares set equal to eight, which gives the step as plus or minus one half.
• Two answers: each sign of the step gives a first term and hence a sixteen-term sum of either seventy-six or twenty.
• Why it matters: expressing symmetric terms as average plus or minus an offset turns the product condition into a clean difference of squares, which is faster and avoids expanding a messy quadratic.

S₁₆ = 76 or 20.

Count the rungs first, then average the ends.

• Use the hint: the hint hands you the rung count, after which the total length is just the count times the average rung length.
• Count the rungs: the spacing divides into the height as ten gaps, so there are eleven rungs in all.
• Average the ends: the average of the longest and shortest rung is thirty-five centimetres.
• Scale to total: eleven rungs times thirty-five centimetres gives three hundred eighty-five centimetres of wood.
• Why it matters: converting all lengths to centimetres up front and reading the sum as count times average length keeps the units consistent and the arithmetic in whole numbers, exactly how the board expects it presented.

385 cm of wood.

See it as the balance point of the whole row.

• Frame it: the chosen house splits the numbers so both sides have an equal sum, which is the balance idea behind the algebra.
• Rearrange the hint: the given condition just says the sum to the chosen house plus the sum before it equals the total of all forty-nine.
• Collapse it: expanding those two sums neatly cancels to the chosen house number squared.
• Solve: so the house number squared equals the total of one thousand two hundred twenty-five, giving thirty-five.
• Why it matters: recognising that the balance condition collapses to a single square is a clean result and shows the house comes out whole only because the total is a perfect square, which earns full reasoning marks.

x = 35.

Factor out the common concrete block.

• The key picture: every step volume is a whole-number multiple of the first step's volume, so treat each step as a stack of identical concrete blocks rather than a separate cuboid.
• Size the block: the unit block is the first step, whose volume is six point two five cubic metres.
• Count the blocks: the fifteen steps hold one, two, three and so on up to fifteen such blocks, which sum to one hundred twenty blocks.
• Multiply back: one hundred twenty blocks at six point two five cubic metres each give a total of seven hundred fifty cubic metres.
• Why it matters: seeing each step as a stack of identical blocks turns the volume problem into the sum of the first fifteen natural numbers, which is quicker and makes the final answer easy to verify.

750 m³.