Key Concepts & Formulae at a Glance
- Arithmetic Progression (AP):A list of numbers where each term is obtained by adding a fixed number d (the common difference) to the preceding term.
- General form:a, a+d, a+2d, a+3d, ... where a is the first term.
- Testing for an AP:a₂−a₁ = a₃−a₂ = ... must hold for every consecutive pair.
- nth term:aₙ = a + (n−1)d
- Sum of first n terms:Sₙ = n2 [2a+(n−1)d] = n2 (a+l), where l is the last term.
- nth term from Sn:aₙ = Sₙ − Sₙ−₁
Exercise 5.1
4 Questions
1In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 14 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
(i) Taxi fare
Fares: 15, 23, 31, 39, ... — each term is 8 more than the last.
Yes, an AP with d = 8.
(ii) Air in cylinder
If the cylinder starts with volume V, removing 14 each time leaves 34 of the previous amount: V, (34)V, (916)V, ... — each term is obtained by multiplying by 34, not by adding a fixed number.
No, not an AP (successive terms don't differ by a constant amount).
(iii) Well digging cost
Costs: 150, 200, 250, 300, ... — each term is 50 more than the last.
Yes, an AP with d = 50.
(iv) Compound interest
Compound interest grows the balance by a fixed percentage each year, not a fixed amount — so the yearly balances multiply by 1.08 each time, rather than differing by a constant.
No, not an AP.
2Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a=10, d=10 (ii) a=−2, d=0 (iii) a=4, d=−3 (iv) a=−1, d=12 (v) a=−1.25, d=−0.25
(i) 10, 20, 30, 40
(ii) −2, −2, −2, −2
(iii) 4, 1, −2, −5
(iv) −1, −12, 0, 12
(v) −1.25, −1.50, −1.75, −2.00
(ii) −2, −2, −2, −2
(iii) 4, 1, −2, −5
(iv) −1, −12, 0, 12
(v) −1.25, −1.50, −1.75, −2.00
3For the following APs, write the first term and the common difference:
(i) 3, 1, −1, −3, ... (ii) −5, −1, 3, 7, ... (iii) 13, 53, 93, 133, ... (iv) 0.6, 1.7, 2.8, 3.9, ...
(i) a = 3, d = −2
(ii) a = −5, d = 4
(iii) a = 13, d = 43
(iv) a = 0.6, d = 1.1
(ii) a = −5, d = 4
(iii) a = 13, d = 43
(iv) a = 0.6, d = 1.1
4Which of the following are APs? If they form an AP, find the common difference d and write three more terms:
(i) 2,4,8,16,... (ii) 2,52,3,72,... (iii) −1.2,−3.2,−5.2,−7.2,... (iv) −10,−6,−2,2,...
(v) 3,3+√2,3+2√2,3+3√2,... (vi) 0.2,0.22,0.222,0.2222,... (vii) 0,−4,−8,−12,... (viii) −12,−12,−12,−12,...
(ix) 1,3,9,27,... (x) a,2a,3a,4a,... (xi) a,a²,a³,a⁴,... (xii) √2,√8,√18,√32,...
(xiii) √3,√6,√9,√12,... (xiv) 1²,3²,5²,7²,... (xv) 1²,5²,7²,73,...
| List | AP? | d / reason | Next three terms |
|---|---|---|---|
| (i) 2,4,8,16 | No | ratios constant, not differences (each term doubles) | — |
| (ii) 2,52,3,72 | Yes | d = 12 | 4, 92, 5 |
| (iii) −1.2,−3.2,−5.2,−7.2 | Yes | d = −2 | −9.2, −11.2, −13.2 |
| (iv) −10,−6,−2,2 | Yes | d = 4 | 6, 10, 14 |
| (v) 3,3+√2,3+2√2,3+3√2 | Yes | d = √2 | 3+4√2, 3+5√2, 3+6√2 |
| (vi) 0.2,0.22,0.222,0.2222 | No | differences 0.02, 0.002, 0.0002 — not constant | — |
| (vii) 0,−4,−8,−12 | Yes | d = −4 | −16, −20, −24 |
| (viii) −12,−12,−12,−12 | Yes | d = 0 | −12, −12, −12 |
| (ix) 1,3,9,27 | No | ratios constant (×3), not differences — this is a GP | — |
| (x) a,2a,3a,4a | Yes | d = a | 5a, 6a, 7a |
| (xi) a,a²,a³,a⁴ | No | a²−a ≠ a³−a² in general (differences depend on a) | — |
| (xii) √2,√8,√18,√32 = √2,2√2,3√2,4√2 | Yes | d = √2 | 5√2, 6√2, 7√2 |
| (xiii) √3,√6,√9,√12 | No | differences √6−√3, 3−√6, 2√3−3 are all different | — |
| (xiv) 1²,3²,5²,7² = 1,9,25,49 | No | differences 8, 16, 24 — not constant | — |
| (xv) 1²,5²,7²,73 = 1,25,49,73 | Yes | d = 24 (the last term is literally 73, not 9²=81, which is what makes this one an AP) | 97, 121, 145 |
Exercise 5.2
20 Questions
1Fill in the blanks in the following table, given that a is the first term, d the common difference and aₙ the nth term of the AP:
| a | d | n | aₙ | |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | 28 |
| (ii) | −18 | 2 | 10 | 0 |
| (iii) | 46 | −3 | 18 | −5 |
| (iv) | −18.9 | 2.5 | 10 | 3.6 |
| (v) | 3.5 | 0 | 105 | 3.5 |
(i) aₙ=a+(n−1)d=7+7(3)=28
(ii) 0=−18+9d → d=2
(iii) −5=a+17(−3) → a=46
(iv) 3.6=−18.9+(n−1)(2.5) → n−1=9 → n=10
(v) d=0 means every term equals a, so aₙ=3.5
(ii) 0=−18+9d → d=2
(iii) −5=a+17(−3) → a=46
(iv) 3.6=−18.9+(n−1)(2.5) → n−1=9 → n=10
(v) d=0 means every term equals a, so aₙ=3.5
2Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, ..., is (A) 97 (B) 77 (C) −77 (D) −87
(ii) 11th term of the AP: −3, −12, 2, ..., is (A) 28 (B) 22 (C) −38 (D) −48½
(i)
a=10, d=−3. a₃₀=10+29(−3)=10−87=−77
(C) −77
(ii)
a=−3, d=−12−(−3)=52. a₁₁=−3+10(52)=−3+25=22
(B) 22
3In the following APs, find the missing terms in the boxes:
(i) 2, □, 26 (ii) □, 13, □, 3 (iii) 5, □, □, 9½ (iv) −4, □, □, □, □, 6 (v) □, 38, □, □, □, −22
(i) 3-term AP: a=2, a₃=26 → d=12 → missing term = 14
(ii) 4-term AP: a₂=13, a₄=3 → d=−5 → a₁=18, a₃=8
(iii) 4-term AP: a₁=5, a₄=9½ → d=32 → a₂=6½, a₃=8
(iv) 6-term AP: a₁=−4, a₆=6 → d=2 → sequence: −4, −2, 0, 2, 4, 6
(v) 6-term AP: a₂=38, a₆=−22 → d=−15 → sequence: 53, 38, 23, 8, −7, −22
(ii) 4-term AP: a₂=13, a₄=3 → d=−5 → a₁=18, a₃=8
(iii) 4-term AP: a₁=5, a₄=9½ → d=32 → a₂=6½, a₃=8
(iv) 6-term AP: a₁=−4, a₆=6 → d=2 → sequence: −4, −2, 0, 2, 4, 6
(v) 6-term AP: a₂=38, a₆=−22 → d=−15 → sequence: 53, 38, 23, 8, −7, −22
4Which term of the AP: 3, 8, 13, 18, ..., is 78?
a=3, d=5. 78=3+(n−1)(5) → n−1=15 → n=16
The 16th term is 78.
5Find the number of terms in each of the following APs:
(i) 7, 13, 19, ..., 205 (ii) 18, 15½, 13, ..., −47
(i) a=7,d=6: 205=7+(n−1)(6) → n−1=33 → n=34
(ii) a=18,d=−2.5: −47=18+(n−1)(−2.5) → n−1=26 → n=27
(ii) a=18,d=−2.5: −47=18+(n−1)(−2.5) → n−1=26 → n=27
(i) 34 terms (ii) 27 terms
6Check whether −150 is a term of the AP: 11, 8, 5, 2 ...
a=11, d=−3. −150=11+(n−1)(−3) → n−1=1613, which is not a whole number.
Since n must be a positive integer, −150 is not a term of this AP.
7Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a₁₆−a₁₁=5d=73−38=35 → d=7. a=a₁₁−10d=38−70=−32.
a₃₁=a+30d=−32+210=178
a₃₁=a+30d=−32+210=178
The 31st term is 178.
8An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
a₃₀−a₃=47d=106−12=94 → d=2. a=a₃−2d=12−4=8.
a₂₉=a+28d=8+56=64
a₂₉=a+28d=8+56=64
The 29th term is 64.
9If the 3rd and 9th terms of an AP are 4 and −8 respectively, which term of this AP is zero?
a₉−a₃=6d=−8−4=−12 → d=−2. a=a₃−2d=4+4=8.
0=8+(n−1)(−2) → n−1=4 → n=5
0=8+(n−1)(−2) → n−1=4 → n=5
The 5th term is zero.
10The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
a₁₄−a₁₀=7d=7 → d=1
d = 1
11Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
a=3, d=12. Need (aₙ−a₄₄)=(n−54)d=132 → n−54=11 → n=65
The 65th term.
12Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
For two APs with the same d, the difference between corresponding terms (aₙ−aₙ') = (a₂−a₁)−(a₂'−a₁') stays constant for every n, since the d's cancel — it always equals the difference of the first terms.
The difference between the 1000th terms is also 100.
13How many three-digit numbers are divisible by 7?
Smallest 3-digit multiple of 7: 105. Largest: 994. AP: 105, 112, ..., 994 with d=7.
994=105+(n−1)(7) → n−1=127 → n=128
994=105+(n−1)(7) → n−1=127 → n=128
128 three-digit numbers are divisible by 7.
14How many multiples of 4 lie between 10 and 250?
First multiple of 4 after 10 is 12; last before 250 is 248. AP: 12, 16, ..., 248 with d=4.
248=12+(n−1)(4) → n−1=59 → n=60
248=12+(n−1)(4) → n−1=59 → n=60
60 multiples of 4 lie between 10 and 250.
15For what value of n, are the nth terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?
AP1: a=63,d=2. AP2: a=3,d=7.
63+(n−1)(2) = 3+(n−1)(7) → 60 = 5(n−1) → n−1=12 → n=13
63+(n−1)(2) = 3+(n−1)(7) → 60 = 5(n−1) → n−1=12 → n=13
n = 13
16Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
a₄−a₃=2d=12 → d=6. a=a₃−2d=16−12=4.
The AP is 4, 10, 16, 22, ...
17Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
a=3,d=5,last=253. Total terms: 253=3+(n−1)(5) → n=51.
20th from the last = (51−20+1)th from the start = 32nd term
a₃₂=3+31(5)=158
20th from the last = (51−20+1)th from the start = 32nd term
a₃₂=3+31(5)=158
The 20th term from the last is 158.
18The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
a₄+a₈=2a+10d=24 → a+5d=12 ... (1)
a₆+a₁₀=2a+14d=44 → a+7d=22 ... (2)
Subtracting: 2d=10 → d=5. From (1): a=12−25=−13.
a₆+a₁₀=2a+14d=44 → a+7d=22 ... (2)
Subtracting: 2d=10 → d=5. From (1): a=12−25=−13.
First three terms: −13, −8, −3
19Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
a=5000,d=200. 7000=5000+(n−1)(200) → n−1=10 → n=11.
n=1 corresponds to 1995, so n=11 corresponds to 1995+10.
n=1 corresponds to 1995, so n=11 corresponds to 1995+10.
His income reached ₹7000 in the year 2005.
20Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
a=5, d=1.75. 20.75=5+(n−1)(1.75) → n−1=9 → n=10
n = 10
Exercise 5.3
20 Questions
1Find the sum of the following APs:
(i) 2,7,12,..., to 10 terms (ii) −37,−33,−29,..., to 12 terms
(iii) 0.6,1.7,2.8,..., to 100 terms (iv) 115,112,110,..., to 11 terms
(i) a=2,d=5,n=10: S=5[4+45]=245
(ii) a=−37,d=4,n=12: S=6[−74+44]=−180
(iii) a=0.6,d=1.1,n=100: S=50[1.2+108.9]=5505
(iv) a=115,d=160,n=11: S=(112)[215+16]=(112)(310)=3320
(ii) a=−37,d=4,n=12: S=6[−74+44]=−180
(iii) a=0.6,d=1.1,n=100: S=50[1.2+108.9]=5505
(iv) a=115,d=160,n=11: S=(112)[215+16]=(112)(310)=3320
2Find the sums given below:
(i) 7+10½+14+...+84 (ii) 34+32+30+...+10 (iii) −5+(−8)+(−11)+...+(−230)
(i) a=7,d=3.5,l=84: n=23, S=(232)(91)=1046.5
(ii) a=34,d=−2,l=10: n=13, S=(132)(44)=286
(iii) a=−5,d=−3,l=−230: n=76, S=(762)(−235)=−8930
(ii) a=34,d=−2,l=10: n=13, S=(132)(44)=286
(iii) a=−5,d=−3,l=−230: n=76, S=(762)(−235)=−8930
3In an AP, find the unknowns:
(i) a=5,d=3,aₙ=50 → n, Sₙ (ii) a=7,a₁₃=35 → d, S₁₃ (iii) a₁₂=37,d=3 → a, S₁₂
(iv) a₃=15,S₁₀=125 → d, a₁₀ (v) d=5,S₉=75 → a, a₉ (vi) a=2,d=8,Sₙ=90 → n, aₙ
(vii) a=8,aₙ=62,Sₙ=210 → n, d (viii) aₙ=4,d=2,Sₙ=−14 → n, a (ix) a=3,n=8,S=192 → d (x) l=28,S=144,n=9 → a
| Given | Find | Answer |
|---|---|---|
| (i) a=5,d=3,aₙ=50 | n, Sₙ | n=16, Sₙ=440 |
| (ii) a=7,a₁₃=35 | d, S₁₃ | d=73, S₁₃=273 |
| (iii) a₁₂=37,d=3 | a, S₁₂ | a=4, S₁₂=246 |
| (iv) a₃=15,S₁₀=125 | d, a₁₀ | d=−1, a₁₀=8 |
| (v) d=5,S₉=75 | a, a₉ | a=−353, a₉=853 |
| (vi) a=2,d=8,Sₙ=90 | n, aₙ | n=5, aₙ=34 |
| (vii) a=8,aₙ=62,Sₙ=210 | n, d | n=6, d=545 |
| (viii) aₙ=4,d=2,Sₙ=−14 | n, a | n=7, a=−8 |
| (ix) a=3,n=8,S=192 | d | d=6 |
| (x) l=28,S=144,n=9 | a | a=4 |
Method note: use Sₙ=(n2)[2a+(n−1)d] and aₙ=a+(n−1)d together, solving whichever two equations match the givenunknown quantities. E.g. (vii): Sₙ=(n2)(a+aₙ)=(n2)(70)=210 → n=6; then aₙ=a+(n−1)d → 62=8+5d → d=545.
4How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
a=9,d=8. Sₙ=(n2)[18+8(n−1)]=636. Solving the resulting quadratic gives n=12 (rejecting the negative root).
n = 12
5The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
S=(n2)(a+l) → 400=(n2)(50) → n=16.
a₁₆=45=5+15d → d=83
a₁₆=45=5+15d → d=83
n = 16, d = 83
6The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
350=17+(n−1)(9) → n−1=37 → n=38.
S=(382)(17+350)=19(367)=6973
S=(382)(17+350)=19(367)=6973
38 terms, sum = 6973
7Find the sum of first 22 terms of an AP in which d=7 and 22nd term is 149.
149=a+21(7) → a=2.
S₂₂=(222)(2+149)=11(151)=1661
S₂₂=(222)(2+149)=11(151)=1661
S₂₂ = 1661
8Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
d=18−14=4, a=14−4=10.
S₅₁=(512)[20+50(4)]=(512)(220)=5610
S₅₁=(512)[20+50(4)]=(512)(220)=5610
S₅₁ = 5610
9If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
S₄=(72)[2a+6d]=49 → a+3d=7
S₁₄=(172)[2a+16d]=289 → a+8d=17
Solving: d=2, a=1.
Sₙ=(n2)[2(1)+(n−1)(2)]=(n2)(2n)=n²
S₁₄=(172)[2a+16d]=289 → a+8d=17
Solving: d=2, a=1.
Sₙ=(n2)[2(1)+(n−1)(2)]=(n2)(2n)=n²
Sₙ = n²
10Show that a₁, a₂, ..., aₙ, ... form an AP where aₙ is defined as below, and find the sum of the first 15 terms in each case:
(i) aₙ=3+4n (ii) aₙ=9−5n
(i) aₙ=3+4n
a₁=7, a₂=11, a₃=15, ... Each term is 4 more than the previous — a genuine AP with a=7, d=4.
S₁₅=(152)[14+14(4)]=(152)(70)=525
S₁₅=(152)[14+14(4)]=(152)(70)=525
AP confirmed, d=4. S₁₅=525
(ii) aₙ=9−5n
a₁=4, a₂=−1, a₃=−6, ... Each term is 5 less than the previous — an AP with a=4, d=−5.
S₁₅=(152)[8+14(−5)]=(152)(−62)=−465
S₁₅=(152)[8+14(−5)]=(152)(−62)=−465
AP confirmed, d=−5. S₁₅=−465
11If the sum of the first n terms of an AP is 4n−n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Sₙ=4n−n².
S₁=4(1)−1²=3, so a₁=3.
S₂=4(2)−2²=4.
a₂=S₂−S₁=4−3=1.
S₃=4(3)−3²=3, so a₃=S₃−S₂=3−4=−1.
S₁₀=4(10)−100=−60, S₉=4(9)−81=−45, so a₁₀=−60−(−45)=−15.
General term: aₙ=Sₙ−Sₙ−₁=(4n−n²)−(4(n−1)−(n−1)²)=5−2n
S₁=4(1)−1²=3, so a₁=3.
S₂=4(2)−2²=4.
a₂=S₂−S₁=4−3=1.
S₃=4(3)−3²=3, so a₃=S₃−S₂=3−4=−1.
S₁₀=4(10)−100=−60, S₉=4(9)−81=−45, so a₁₀=−60−(−45)=−15.
General term: aₙ=Sₙ−Sₙ−₁=(4n−n²)−(4(n−1)−(n−1)²)=5−2n
S₁=3, S₂=4, a₂=1, a₃=−1, a₁₀=−15, aₙ=5−2n
12Find the sum of the first 40 positive integers divisible by 6.
AP: 6,12,18,...; a=6,d=6,n=40.
S₄₀=(402)[12+39(6)]=20(246)=4920
S₄₀=(402)[12+39(6)]=20(246)=4920
4920
13Find the sum of the first 15 multiples of 8.
AP: 8,16,24,...; a=8,d=8,n=15.
S₁₅=(152)[16+14(8)]=(152)(128)=960
S₁₅=(152)[16+14(8)]=(152)(128)=960
960
14Find the sum of the odd numbers between 0 and 50.
AP: 1,3,5,...,49; a=1,d=2,l=49. n=49−12+1=25.
S=(252)(1+49)=25(25)=625
S=(252)(1+49)=25(25)=625
625
15A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days?
a=200,d=50,n=30.
S₃₀=(302)[400+29(50)]=15(1850)=27750
S₃₀=(302)[400+29(50)]=15(1850)=27750
₹27,750
16A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
n=7,d=−20,S=700. S=(72)[2a−120]=700 → 2a−120=200 → a=160.
Prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40
17In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Trees per class (one section) = class number, so total across 1 section of every class = 1+2+...+12 = 12×132 = 78.
With 3 sections per class: total = 3 × 78 = 234
With 3 sections per class: total = 3 × 78 = 234
234 trees will be planted.
18A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π=227)
Each semicircle's length = πr. The radii 0.5, 1.0, 1.5, ... form an AP with a=0.5, d=0.5, n=13.
Sum of radii = (132)[1+12(0.5)]=(132)(7)=45.5
Total length = π × 45.5 = (227) × 45.5 = 143
Sum of radii = (132)[1+12(0.5)]=(132)(7)=45.5
Total length = π × 45.5 = (227) × 45.5 = 143
143 cm
19200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
a=20,d=−1,Sₙ=200. Solving (n2)[40−(n−1)]=200 gives two roots, n=16 or n=25.
At n=25, the term a₂₅=20−24=−4, which is impossible (can't have a negative number of logs in a row), so this root is rejected.
At n=16: a₁₆=20−15=5
At n=25, the term a₂₅=20−24=−4, which is impossible (can't have a negative number of logs in a row), so this root is rejected.
At n=16: a₁₆=20−15=5
16 rows, with 5 logs in the top row.
20In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Distances of the potatoes from the bucket: 5, 8, 11, ..., (10 terms, an AP with a=5, d=3).
Each potato requires a round trip, so the total distance is twice the sum of these distances.
Sum of distances: S₁₀=(102)[10+9(3)]=5(37)=185
Total distance = 2 × 185 = 370
Each potato requires a round trip, so the total distance is twice the sum of these distances.
Sum of distances: S₁₀=(102)[10+9(3)]=5(37)=185
Total distance = 2 × 185 = 370
The competitor runs a total of 370 m.
Exercise 5.4 (Optional)
5 Questions
1Which term of the AP: 121, 117, 113, ..., is its first negative term?
a=121, d=−4. Need aₙ<0: 121+(n−1)(−4)<0 → n>1254=31.25
The smallest integer n satisfying this is n=32. Check: a₃₁=121+30(−4)=1 (still positive), a₃₂=121+31(−4)=−3 (negative).
The smallest integer n satisfying this is n=32. Check: a₃₁=121+30(−4)=1 (still positive), a₃₂=121+31(−4)=−3 (negative).
The 32nd term (value −3) is the first negative term.
2The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Let a₃=x, a₄=6−x. Product: x(6−x)=8 → x²−6x+8=0 → x=2 or x=4.
Case 1: a₃=2, a₄=4 → d=4−24=12, a=a₃−2d=2−1=1. S₁₆=(162)[2(1)+15(12)]=8(9.5)=76
Case 2: a₃=4, a₄=2 → d=2−44=−12, a=4−2(−12)=5. S₁₆=(162)[2(5)+15(−12)]=8(2.5)=20
Case 1: a₃=2, a₄=4 → d=4−24=12, a=a₃−2d=2−1=1. S₁₆=(162)[2(1)+15(12)]=8(9.5)=76
Case 2: a₃=4, a₄=2 → d=2−44=−12, a=4−2(−12)=5. S₁₆=(162)[2(5)+15(−12)]=8(2.5)=20
S₁₆ = 76 or 20 (both cases are valid, giving two different possible APs).
3A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs?
Number of rungs = 25025 + 1 = 11 (using the hint).
Rung lengths form an AP: a=45, l=25, n=11.
Total wood = S = (112)(45+25) = (112)(70) = 385
Rung lengths form an AP: a=45, l=25, n=11.
Total wood = S = (112)(45+25) = (112)(70) = 385
385 cm of wood is required.
4The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Sum of house numbers 1 to (x−1) = (x−1)x2.
Sum of house numbers (x+1) to 49 = S₄₉ − Sₓ = 49502 − xx+12.
Setting these equal: (x−1)x2 = 1225 − xx+12
x(x−1) + x(x+1) = 2450 → 2x² = 2450 → x² = 1225 → x = 35
Sum of house numbers (x+1) to 49 = S₄₉ − Sₓ = 49502 − xx+12.
Setting these equal: (x−1)x2 = 1225 − xx+12
x(x−1) + x(x+1) = 2450 → 2x² = 2450 → x² = 1225 → x = 35
x = 35 (house number 35 has equal sums of house numbers on either side).
5A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of ¼ m and a tread of ½ m. Calculate the total volume of concrete required to build the terrace.
Using the hint, the volume of the k-th step (from the bottom) = k × (¼ × ½ × 50) = k × 6.25 m³ — since each step's height is k times the rise, over the same tread and length.
Total volume = 6.25 × (1+2+...+15) = 6.25 × (15×162) = 6.25 × 120 = 750
Total volume = 6.25 × (1+2+...+15) = 6.25 × (15×162) = 6.25 × 120 = 750
750 m³ of concrete is required.
Frequently Asked Questions
Each term after the first is obtained by adding a fixed number (the common difference, d) to the term before it. Check by confirming a₂−a₁ = a₃−a₂ = a₄−a₃ = ... is the same value throughout; if even one pair of consecutive differences disagrees, it isn't an AP.
Use aₙ = a + (n−1)d, where a is the first term and d is the common difference. This lets you jump straight to any term without listing every term in between.
Sₙ = n2 [2a + (n−1)d], or equivalently Sₙ = n2 (a + l) when the last term l is known instead of d. Both forms give the same result.
This happens when the common difference is negative and the first term is positive (or vice versa) — some terms are positive and some are negative, so a block of terms in the middle can sum to zero. Both term counts that satisfy the equation are then mathematically valid.
Yes — a printable Formula Card covering all 7 units, including Arithmetic Progressions, is available for ₹99.
Continue to the Next Chapter
Chapter 6 (Triangles) moves from number patterns into geometry.