NCERT Exemplar Class 10 Maths Chapter 5 Exercise 5.3 Solutions

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.