NCERT Exemplar Solutions Class 10 Maths Chapter 5

Walk backwards from the seventh term.

• Count the steps: going from the first term to the seventh adds the common difference six times, so the total shift is 6×(-4)=-24.
• Undo the shift: since the seventh term is 4, the start must have been 24 higher, that is 4+24=28.
• Check forward: starting at 28 and stepping down by 4 gives 28,24,20,16,12,8,4, and the seventh entry is indeed 4, so the first term is 28.

Option (D), a=28.

Zero step means no change.

• What d does: the common difference is the size of each jump from one term to the next, and here that jump is 0.
• Effect on the list: with a zero jump the list never moves off its starting value of 3.5, so the 2nd term and the 101st term are the same 3.5.
• Spot the trap: the large value of n is only a distractor, because multiplying it by d=0 wipes it out completely.

Option (B), 3.5.

Test the gaps, not the terms.

• The deciding rule: an AP is decided purely by whether the gaps between neighbours stay constant.
• Apply it: stepping from -10 to -6 is a rise of 4, and the same rise of 4 repeats from -6 to -2 and from -2 to 2, so the list is an AP with common difference 4.
• The trap options: option (A) comes from a sign error and option (C) from reversing the order of subtraction.

Option (B), d=4.

Ten steps of half-fives.

• Size of each step: every move up the list adds 52, and reaching the 11th term takes 10 such moves, a total of 10×52=25.
• Add to the start: putting that on the first term -5 gives -5+25=20.
• Pattern check: every two steps add a whole 5, so ten steps add 25, which lands the eleventh term at 20 as well.

Option (B), 20.

Keep stepping down by two.

• Build the list: from -2, each new term is two less than the one before, giving -2,-4,-6,-8.
• Why the traps fail: options (B) and (D) multiply instead of add, jumping -24→-8 or doubling to -16, which breaks the constant-difference rule of an AP.
• The clean route: only the steady step of -2 keeps the gap fixed and produces the correct list.

Option (C), -2,-4,-6,-8.

One subtraction, then one formula.

• Find the step: the jump from -3 to 4 is a rise of 7, so d=7.
• Reach term 21: the twenty-first term sits 20 jumps past the start, adding 207=140 to -3 for a result of 137.
• Rule out the trap: the negative option -143 tries to trick you into subtracting, but the terms clearly increase once d=7 is positive.

Option (B), 137.

Use the gap between known terms.

• Find the step: going from the 2nd term to the 5th covers 3 steps and a rise of 25-13=12, so each step is d=12/3=4.
• Extend to term 7: the seventh term is two steps beyond the fifth, adding 24=8 to 25 to give 33.
• Why it is faster: this term-gap shortcut avoids solving for the first term a at all, yet lands on the same answer.

Option (B), 33.

The list is just the 21 times table.

• Spot the pattern: every term is a multiple of 21, the first being 211 and the second 212, so the nth term is simply 21n.
• Solve directly: setting 21n=210 gives n=10 at once.
• Why it works: the common difference happens to equal the first term, which turns the AP into a clean multiplication table and beats the full formula here.

Option (B), the 10th term.

Count the steps between the two terms.

• Count the steps: from the 13th term to the 18th term there are 18-13=5 steps, each worth the common difference 5.
• Total the rise: so the climb is 55=25, regardless of where the AP actually starts.
• Why no a is given: the first term never enters the answer, which is exactly why the question supplies no value for it.

Option (C), 25.

Divide the rise by the steps.

• Per-step size: over 4 steps the terms climb by 32, so a single step is 32/4=8.
• Read off d: that single-step size is exactly the common difference, giving d=8.
• Reject negatives: the negative choices would only fit a decreasing AP, but a positive total rise of 32 rules them out.

Option (A), d=8.

The shared d does nothing to the gap.

• What cancels: both fourth terms carry the same 3d, so subtracting one from the other makes the 3d vanish.
• What survives: only the starting gap is left, and that gap is -1-(-8)=7.
• Why d is hidden: the step is the same in both APs and cancels for every matching pair of terms, not just the fourth, so its value is never asked for.

Option (C), 7.

Solve the relation, then jump to term 18.

• Simplify: expanding 7(a+6d)=11(a+10d) leads to -4a=68d, so a=-17d.
• Evaluate term 18: the eighteenth term is a+17d, and replacing a gives -17d+17d=0.
• No accident: the condition 7a₇=11a₁₁ is built so the position 7+11=18 is forced to zero whatever the value of d.

Option (D), 0.

Read the AP from right to left.

• Step from the end: looking back from 49, each step down is the same d=3, and the first-from-end is 49 itself.
• Count back three: the fourth-from-end is three steps back, 49-33=40.
• Reverse and read: listing the AP backwards as 49,46,43,40, shows the fourth entry as 40 too, without ever finding how many terms the AP has.

Option (B), 40.

Match the name to the method.

• The famous trick: the clever pairing of 1 with 100, 2 with 99 and so on into fifty 101s is the schoolboy story of Carl Friedrich Gauss, giving 50101=5050.
• Rule out the rest: Pythagoras and Euclid are linked with geometry and Newton with calculus, none of them with this sum trick.
• Tie to the formula: the same pairing is the reason the first-plus-last sum formula works, so the correct name is Gauss.

Option (C), Gauss.

The terms cancel in pairs.

• List them: the first six terms are -5,-3,-1,1,3,5.
• Pair the outers: -5+5=0, -3+3=0 and -1+1=0, so the whole sum is 0.
• Formula agrees: the sum formula gives 3[2(-5)+5(2)]=3[-10+10]=0, and the symmetry around zero is what makes the total vanish exactly.

Option (A), 0.

A falling AP drags the sum negative.

• Direction first: the terms start at 10 and drop by 4 each time, so they soon go negative and pull the total below zero.
• Plug in: the sum formula gives 8[20-60]=8(-40)=-320.
• Sift the options: the size 320 with a minus sign rules out every positive choice, while the wrong value -352 comes from miscounting the steps as 16 instead of 15.

Option (A), -320.

Use first-plus-last directly.

• Average the ends: both ends of the AP are known, 1 and 20, so each pair averages 1+202=10.5.
• Divide to count: the sum 399 divided by this average gives the number of terms, 399/10.5=38, the same as n=2399/21=38.
• Skip d: choosing this formula avoids solving for the common difference, which the question never actually needs.

Option (C), 38.

Add the small list straight.

• Direct addition: the five numbers are 3,6,9,12,15, and adding them gives 3+6+9+12+15=45.
• Factor instead: pulling the 3 out gives 3(1+2+3+4+5)=315=45, the same total.
• Formula check: the sum formula confirms 52(6+12)=45, though with only five terms direct addition is the quickest safe route.

Option (A), 45.

One test, applied seven times.

• The single rule: the only question to ask of each list is whether the gaps between neighbours stay equal all the way along, and nothing else matters.
• Clear passes: lists (i) and (iv) have flat, unchanging gaps of 0 and 11, so they are arithmetic progressions straight away with no extra work.
• Hidden pass: list (vii) buries its AP behind surds, but once it is rewritten as 3,23,33,43 the gap is a steady 3, so it also qualifies.
• The failures: the rest each break the rule in their own way, since (ii) and (iii) bounce up and down, list (v) shrinks towards zero, and list (vi) keeps doubling, so none of them holds a constant gap.
• Verdict: only constant gaps count, which leaves (i), (iv) and (vii) as the genuine arithmetic progressions here.

APs: (i), (iv), (vii). Not APs: (ii), (iii), (v), (vi).

The fourth term betrays the list.

• The bait: the first two gaps are both -12, which is exactly what the claim leans on.
• The giveaway: the fourth term 52 jumps up by 92 from -2, nothing like -12, so the gap changes and the list fails the AP test.
• The lesson: matching only a₂-a₁=a₃-a₂ is too weak, since a genuine AP needs the same gap running all the way through.

False; the gap is not constant (a₄-a₃=92).

The starting point cancels out.

• Write both terms: a₃₀=a+29d and a₂₀=a+19d both carry the same first term a.
• Subtract: the a disappears and leaves (29-19)d=10d, which with d=-4 is -40.
• Cross-check: computing the two large terms, -119 and -79, and subtracting gives the same -40, but the gap rule reaches it in a single line.

Yes; the answer is 10d=-40.

Equal steps keep the distance fixed.

• Lockstep: both APs add the same d at each move, so they rise together at exactly the same rate.
• What cancels: whatever term number you pick, the (n-1)d contributions are identical and subtract away.
• What is left: only the unchanging head start of 7-2=5 survives, so the 10th-term gap, the 21st-term gap and every corresponding-term gap are all the same 5.

All differences equal 5, the gap between the first terms.

The terms step over zero.

• List it out: starting at 31 and dropping by 3 gives 31,28,25,22,,4,1,-2,, which lands on 1 and then -2, jumping straight past 0.
• Confirm by algebra: demanding a_n=0 forces n=343, a non-integer position, so the value is never actually reached.
• Verdict: since n must be a whole number, 0 is not a term of this AP.

No; the terms go ,4,1,-2,, skipping 0.

Add as the meter ticks.

• Running total: a real taxi meter keeps a running total of 15, then 23, then 31, then 39, each 8 more than the last, which is a textbook AP.
• The misread: the statement's list 15,8,8,8 is the charge for each individual km and never accumulates, so it confuses ``per km'' with ``total''.
• Verdict: read correctly, the cumulative fares do form an AP, so the statement is false.

False; running totals 15,23,31,39, are an AP, d=8.

Look for a steady add-on.

• Flat fee: a repeating Rs 400 fee has gap 0, which still counts as an AP.
• Fixed climbs: class fees climb by a fixed Rs 50 and a simple-interest balance climbs by a fixed Rs 100, both clean arithmetic progressions.
• The odd one out: bacteria double, so the increase itself grows from a to 2a to 4a, a multiplying pattern that is a GP and not an AP, leaving only the first three as APs.

APs: (i), (ii), (iii). Not an AP: (iv).

Degree of n decides it.

• The signature: an AP's nth term is a straight-line function of the form dn+(a-d), so a valid formula must be degree one in n and nothing higher.
• The clean fit: the expression 2n-3 matches that shape exactly, with slope d=2, so the first list is a genuine AP.
• Why the squares fail: the other two carry an n² term, and working out their gaps gives 6n-3 and 2n, both of which keep growing as n grows.
• Verdict: a changing gap is forbidden in an AP, so only the linear form (i) qualifies and the two quadratic formulas do not.

Only (i) is an AP nth term (d=2).

Four short calculations, one per row.

• Listed terms: row A₁ shows terms, so d is a direct subtraction giving -4.
• Distant terms: rows A₂ and A₃ give a far-off term, so divide its gap from the first term by the step count, 18/9=2 and 6/9=23.
• Two apart: row A₄ gives two terms two steps apart, so halve their gap, (3-13)/2=-5.
• Match up: lining these against Column B gives B₄, B₅, B₁ and B₂ in turn.

A₁B₄, A₂B₅, A₃B₁, A₄B₂.

Find d, then keep adding it.

• Verify first: each list passes the constant-gap test with d=14,-13,3,1,a+1 in turn.
• Extend mechanically: once d is known, the next three terms come from adding it three more times.
• Letters are harmless: the symbolic lists (iv) and (v) look fearsome but behave like any AP, since the gaps 1 and a+1 stay fixed regardless of the symbols, so the same add-on rule finishes them.

Next three terms found by adding the verified d thrice in each part (see boxed list above).

Three plug-ins, nothing more.

• The recipe: the first three terms are simply a, then a+d, then a+2d.
• Part by part: the sixths in (i) combine to 12,13,16, and the steps of -3 in (ii) give -5,-8,-11.
• Surd part: in (iii), writing everything over 2 turns 2+12 into 32 and the next into 42, with no rule beyond adding d each time.

First three terms as listed in the box above.

Anchor on the two known terms.

• Fix the step: the terms 7 at position 2 and 23 at position 4 sit two steps apart, so 2d=16 and d=8.
• Fill the gaps: a is one step below 7 giving -1, b is one step above 7 giving 15, and c is one step above 23 giving 31.
• Check: the full list -1,7,15,23,31 rises by a steady 8, as it should.

a=-1, b=15, c=31.

The difference clue isolates d.

• Get d free: subtracting the 8th term from the 13th leaves 5d=20, so d=4 at once with no a involved.
• Then get a: feeding d=4 into a₅=a+16=19 gives a=3, so the AP starts at 3 and climbs by 4 as 3,7,11,15,
• Order matters: starting from the gap clue keeps the algebra cleaner than tackling a₅ first.

3,7,11,15,…

Build the AP, then count to the end.

• Shape the AP: the gap between the 26th and 11th terms is 15d=0-3=-3, so d=-15, and then a₁₁=a-2=3 gives a=5.
• Count the length: the last term -15 must satisfy 5+(n-1)(-15)=-15, which unwinds to n-1=26, so there are 27 terms.
• One job each: two clues shape the AP and the third counts how far it runs.

d=-15, n=27.

Two linear clues, one subtraction.

• Use symmetry: the fifth and seventh terms straddle the sixth, so their sum 52 means a₆=26.
• Find the step: with a₁₀=46, the four-step gap from the 6th to the 10th term is 46-26=20, giving d=5.
• Back to the start: then a=a₆-5d=26-25=1, so the AP is 1,6,11,16, and the symmetry trick made the first clue almost free.

1,6,11,16,…

Decode the wording, then march to term 20.

• Read the phrase: ``24 less'' fixes the four-step gap a₁₁-a₇=4d=24, so d=6.
• March forward: starting from the given a=12 and adding 19 steps of 6 gives 12+114=126.
• The only trap: the wording is the hard part, but once it is read as a positive gap of 24 over four steps the arithmetic is straightforward.

a₂₀=126.

Measure both terms from the zero.

• Step from zero: since the 9th term is 0, counting forward 20 steps reaches the 29th term at 20d and 10 steps reaches the 19th term at 10d.
• The doubling: because 20d is exactly twice 10d, the 29th term is twice the 19th.
• Why it is clean: the zero term acts like a fresh origin, making the doubling obvious from the step counts alone.

a₂₉=2 a₁₉.

Test membership through the position.

• Term rule: the AP climbs by 3 from 7, so any term is 7+3(n-1).
• Solve for n: demanding this equals 55 gives 3(n-1)=48 and n=17, a clean whole number, so 55 is genuinely the 17th term.
• The test: had n come out fractional, the value would have fallen between two terms and not belonged to the AP at all.

Yes; the 17th term.

Force the two gaps to agree.

• Find the gaps: subtracting in pairs gives the left gap as k²-k-2 and the right gap as k²+k-2.
• Set them equal: matching the gaps removes the k² and the constant, leaving -k=k, so k=0.
• Why it collapses: the quadratic clutter looks intimidating but falls apart because the squared terms are identical on both sides.

k=0.

Centre on the middle part.

• Symmetric form: calling the parts a-d,a,a+d turns the sum 207 into 3a, so the middle part is a=69 outright.
• Use the product: the two smaller parts are 69-d and 69, and their product 4623 gives 69-d=67, hence d=2.
• Confirm: the three numbers are 67,69,71, which add to 207 and have 6769=4623 as required.

67, 69, 71.

Symmetry pins the middle, the ratio sets the spread.

• Lock the middle: three angles in AP sum to 3a=180^∘, so the middle angle is fixed at 60^∘ straight away, before the second condition is even used.
• Set the spread: the condition that the greatest angle is twice the least becomes 60+d=2(60-d), which solves to give the common difference d=20^∘.
• Read the angles: the three angles then spread out to 40^∘, 60^∘ and 80^∘, sitting evenly around the locked middle of 60^∘.
• Both checks pass: the three angles add back to 180^∘ and the largest is indeed twice the smallest since 80=240, so the answer is consistent on both counts.

40^∘, 60^∘, 80^∘.

Equate two formulas, solve once.

• Two term rules: the first list shrinks by 2 giving 11-2n, and the second shrinks by 3 giving 27-3n.
• Set equal: they agree when 11-2n=27-3n, which rearranges to n=16.
• Find the value: plugging back gives the shared value 11-32=-21, so the two lists, though starting far apart at 9 and 24, cross at their 16th terms.

n=16, common term =-21.

Two paired sums, one clean subtraction.

• Same head: each clue doubles up the first term, so both read 2a plus a multiple of d.
• Eliminate: subtracting them wipes out the 2a and gives 10d=-10, hence d=-1 and then a=8.
• Finish: the tenth term is 8+9(-1)=-1, and the two-term structure of the clues is exactly what makes the elimination so quick.

a₁₀=-1.

Count back from the last term.

• Reverse the step: the AP ends at -100 and steps by -2, so reading it backwards means adding 2 each step.
• Eleven steps back: the 12th-from-end is 11 steps back from -100, that is -100+112=-78.
• Or just flip it: reversing the AP into -100,-98,-96, and taking the 12th entry again gives -78.

-78.

Cross from positive to negative.

• Stay positive: the terms fall by 5 from 53, so they remain positive while 58-5n>0, that is up to n=11 with value 3.
• First dip: the very next term at n=12 drops to -2, the first one below zero.
• Inequality route: solving n>11.6 and rounding up to 12 gives the same answer without listing every term.

The 12th term, a₁₂=-2.

List the endpoints, then count.

• The pattern: remainder-3 numbers run ,11,15,19,,299,, stepping by 4.
• Pin the ends: the first one inside the range is 11 and the last is 299.
• Count the AP: from 11 to 299 in steps of 4 there are 2884+1=73 numbers, with the endpoints doing the heavy lifting and the +1 finishing the count.

73 numbers.

Count first, then grab the centre pair.

• Find the count: solving 133=-43+(n-1)13 gives n=18, an even number of terms.
• Grab the centre: the two middle terms are the 9th and 10th, 43 and 53, whose sum is 3.
• Neat check: in any AP the two middle terms average the first and last, and 12(-43+133)=32, so their sum is 3, matching.

Sum =3.

Get the count, then the step.

• Count from the average: the first-plus-last sum is -5+45=40, so each term averages 20, and dividing the total 120 by that average gives n=6 terms.
• Then the step: with six terms spanning -5 to 45 there are five gaps, so d=45-(-5)5=10.
• Two formulas: the sum formula hands over n and the last-term formula then hands over d.

n=6, d=10.

Three sums, three tactics.

• Numeric AP: part (i) finds n=80 from the last term -236, then 802(1-236)=-9400.
• Split the sum: part (ii) breaks as 4n minus 1n times the triangular number n(n+1)2, simplifying to 7n-12.
• Factor the denominator: part (iii) keeps the fixed denominator a+b aside and sums the numerator AP, first term a-b and d=2a-b, over 11 terms to 11(11a-6b).
• The key: each part hides a plain AP inside a disguise, and spotting it is what makes all three sums routine.

(i) -9400; (ii) 7n-12; (iii) 11(11a-6b)a+b.

Two stages: locate, then add.

• Locate the term: the AP drops by 5 from -2, so -77 sits where -5(n-1)=-75, that is n=16.
• Add the terms: averaging the ends gives -2+(-77)2=-39.5, and times 16 that is -632, the same as 162(-79)=-632.
• Order matters: pinpointing n=16 first is what makes the sum straightforward.

16th term; sum =-632.

Constant gap proves it, then sum.

• Prove the AP: subtracting consecutive terms of 3-4n always gives -4, so the list is an AP starting at -1 with step -4.
• Sum twenty: the bracket is 2(-1)+19(-4)=-78, and half of 20 times that is -780.
• The shortcut: the linear shape of a_n guarantees the AP, and the standard sum formula does the rest.

AP with d=-4; S₂₀=-780.

Differences of sums give the terms.

• First term: the first sum is S₁=5, so a₁=5.
• Peel off more: the next sums 18 and 39 give a₂=18-5=13 and a₃=39-18=21, a steady step of 8, so the AP is 5,13,21,
• Formula check: the direct route a_n=S_n-S_n-1=8n-3 gives the same terms, confirming a=5 and d=8.

5,13,21,…

Turn the sum into a term rule.

• Build the rule: subtracting S_n-1 from S_n=3n²+5n collapses the squares to give the clean linear term a_n=6n+2.
• Solve for k: demanding 6k+2=164 yields 6k=162 and k=27.
• Why this tool: the quadratic sum hides a simple linear term, which is exactly why a_n=S_n-S_n-1 is the right approach.

k=27.

Reduce both sides to a and d.

• Write the sums: the three sums expand to S₁₂=12a+66d, S₈=8a+28d and S₄=4a+6d.
• Form the right side: the difference S₈-S₄=4a+22d, tripled, is 12a+66d, which is exactly S₁₂.
• The conclusion: both sides reduce to the same expression in a and d, so the identity holds for every AP whatever its first term and step.

S₁₂=3(S₈-S₄).

Two terms set up the whole sum.

• Get the step: the drop from the 4th to the 9th term, -15 over five steps, gives d=-3.
• Get the start: then a₄=-15 yields a=-6.
• Sum it: the bracket for seventeen terms is 2(-6)+16(-3)=-60, and half of 17 times that is -510, just one substitution once a and d are known.

S₁₇=-510.

Solve the simultaneous pair.

• Strip constants: S₆ and S₁₆ reduce to 2a+5d=12 and 2a+15d=32.
• Eliminate: subtracting kills the 2a and leaves 10d=20, so d=2 and a=1.
• Finish: the ten-term sum is then 5(2+18)=100, and forcing each sum into the same 2a-plus form is what makes the elimination immediate.

S₁₀=100.

The middle term carries the average.

• Symmetry: in any AP the terms are symmetric about the centre, so the middle term is the mean of them all.
• Multiply: with eleven terms and a middle of 30, the total is simply 1130=330.
• Why it fits: this shortcut sidesteps finding a and d entirely, which the question never supplies.

S₁₁=330.

Reverse, then sum ten.

• Flip the AP: the last ten terms read 126,124,, a fresh AP with start 126 and step -2.
• Sum them: its sum is 5(252-18)=1170.
• Cross-check: the 10th-from-last term is 126-92=108, and averaging the ends gives 126+108210=1170, the same total.

1170.

Reduce to multiples of 18.

• Use the LCM: being a multiple of 2 and of 9 at once is the same as being a multiple of 18, so the seven numbers are 18,36,,126.
• Sum directly: their sum is 72(36+108)=504.
• Factor instead: pulling 18 out gives 18(1+2+⋯+7)=1828=504, the same total.

504.

A quadratic with two honest roots.

• Form the quadratic: the sum condition becomes n²-16n+55=0, factoring to (n-5)(n-11)=0, so n=5 or 11.
• Why both hold: the terms from the 6th to the 11th, namely -5,-3,-1,1,3,5, cancel to 0.
• The reason: adding those six extra terms leaves the total of -55 unchanged, which is exactly why two answers appear.

n=5 or 11; terms 6 to 11 sum to zero.

Two sum expressions, set equal.

• Write both sums: the first AP's n-term sum is 10n²-2n, while the second AP's 2n-term sum is 16n²-68n.
• Equate: tidying gives 6n²-66n=0, that is 6n(n-11)=0, so discarding the useless n=0 leaves n=11.
• The one trap: the second sum runs to 2n terms, not n, and using that count carefully is the only subtlety.

n=11.

Account for every rupee.

• Sum the savings: the piggy-bank deposits add to 1+2+⋯+31=496 using 31322.
• List the parts: the pocket money had to cover three things, the 496 saved, the 204 spent and the 100 still in hand.
• Add up: summing them gives 496+204+100=800, with the AP handling only the savings and the rest being plain bookkeeping.

Rs 800.

Set total savings to the target.

• Model the savings: saving 32,36,40, is an AP with step 4, and its n-month total is n(2n+30).
• Form the quadratic: setting this to 2000 gives n²+15n-1000=0, which factors as (n-25)(n+40)=0.
• Pick the root: only the positive root n=25 makes sense for a month count, so it takes 25 months to reach Rs 2000.

25 months.

Set up two equations, solve, then extend.

• Two clues: writing the sums out, the first clue becomes 12a+31d=167 and the ten-term clue becomes 2a+9d=47.
• Solve: eliminating a gives d=5, then a=1.
• Extend: with the AP fixed, the twenty-term sum is 10(2+95)=970, so the work is two linear equations followed by one clean plug-in.

S₂₀=970.

Three sums tied together by LCM.

• Two near-twins: parts (i) and (ii) are both multiples of 10, differing only in whether the last term 500 is counted, so 49 terms sum to 12250 and 50 terms sum to 12750.
• Inclusion-exclusion: part (iii) wants the ``or'' count, so add the multiples of 2 giving 62750 and the multiples of 5 giving 25250, then remove the doubly counted multiples of 10.
• Finish: subtracting that overlap of 12750 once leaves 75250, and this overlap step is the heart of the third part.

(i) 12250; (ii) 12750; (iii) 75250.

Translate, clear fractions, solve.

• First clue: doubling the ``eighth is half the second'' sentence clears the half and gives the relation a=-13d.
• Second clue: tripling the ``exceeds one third by one'' sentence clears the third and gives 2a+27d=3.
• Combine: substituting the first relation into the second leaves d=3 and then a=-39, so the fifteenth term is -39+14(3)=3, with each English clause turning into one tidy equation once the denominators are gone.

a₁₅=3.

Collapse each triple to its centre.

• Middle triple: the three central terms average to the 19th, so the sum 225=3a₁₉ gives a₁₉=75 at once.
• Last triple: the same trick on the final three gives a₃₆=143, two clean single-term facts instead of six.
• Solve: the gap a₃₆-a₁₉=17d=68 yields d=4, and back-substitution gives a=3, so the AP is 3,7,11,15, with the triple-to-centre shortcut clearing all the clutter.

3,7,11,15,…

Sum the easy set, then subtract.

• Wanted set: the multiples of 9 between 100 and 200 are eleven terms from 108 to 198, summing to 1683.
• Whole range: every integer from 101 to 199 adds up to 14850, an easy first-plus-last sum.
• Complement: removing the multiples of 9 from that total leaves 14850-1683=13167 for the non-multiples, which is far quicker than summing the 89 scattered non-multiples directly.

(i) 1683; (ii) 13167.

One relation powers both ratios.

• The key relation: the given 2:3 term ratio cross-multiplies and rearranges down to the single fact a=4d.
• Term ratio: substituting makes the fifth and twenty-first terms 8d and 24d, which reduces to the ratio 1:3.
• Sum ratio: likewise the sums become 30d and 294d, a ratio of 5:49, and because d cancels each time the answers come out as pure numbers needing no value of d.

1:3 and 5:49.

Build n, then substitute into the sum.

• Step size: the second term being b fixes the common difference as d=b-a, the only step the proof needs.
• Count the terms: the last-term equation c=a+(n-1)(b-a) unwinds to n=b+c-2ab-a.
• Substitute: feeding this into S_n=n2(a+c) produces the target form straight away, so the proof is just two standard AP formulas chained together with symbols in place of numbers.

S_n=(a+c)(b+c-2a)2(b-a).

Sum to 437, then read off the last term.

• Set up the sum: the series steps by 3 from -4, so its n-term sum is n2(3n-11).
• Solve the quadratic: setting this equal to 437 gives 3n²-11n-874=0, whose only positive root is n=19.
• Read off x: the nineteenth term is -4+183=50, which is the unknown x, so the equation holds exactly when the running sum reaches its 19th term.

x=50 (at n=19).

One term, one sum, one subtraction.

• The 30th payment: the instalments climb by Rs 100 each month, so the thirtieth is 1000+29100=3900.
• Total paid: the sum of all thirty instalments is S₃₀=154900=73500, the money handed over so far.
• Balance left: taking that off the Rs 118000 loan leaves Rs 44500 outstanding, so the term formula answers the first part and the sum formula, minus the total, answers the second.

30th instalment Rs 3900; balance Rs 44500.

Symmetry, round trips, and one AP.

• The layout: the middle 14th flag holds the store, so 13 flags lie on each side at distances 2,4,,26 m.
• Round trips: each flag needs a there-and-back walk of 2x, so one side totals 2(2+4+⋯+26)=2182=364 m and both sides give 728 m.
• Longest carry: the single hardest trip is to the farthest flag at 26 m, with the AP sum 2+4+⋯+26=182 doing the main work and the symmetry and doubling wrapped around it.

Total 728 m; longest single carry 26 m.