NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2

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.