Exercise 5.3 Solutions
Arithmetic Progressions
1. Find the sum of the following APs:
Formula: Sn = n/2 [2a + (n-1)d]
(i) 2, 7, 12, …, to 10 terms.
a=2, d=5, n=10. S10 = 10/2 [2(2) + (10-1)5] = 5[4 + 45] = 5(49) = 245.
(ii) -37, -33, -29, …, to 12 terms.
a=-37, d=4, n=12. S12 = 12/2 [2(-37) + (12-1)4] = 6[-74 + 44] = 6(-30) = -180.
(iii) 0.6, 1.7, 2.8, …, to 100 terms.
a=0.6, d=1.1, n=100. S100 = 100/2 [2(0.6) + (100-1)1.1] = 50[1.2 + 108.9] = 50(110.1) = 5505.
(iv) 1/15, 1/12, 1/10, …, to 11 terms.
a=1/15, d=1/12 – 1/15 = 1/60, n=11. S11 = 11/2 [2(1/15) + (11-1)(1/60)] = 11/2 [2/15 + 10/60] = 11/2 [2/15 + 1/6] = 11/2 [9/30] = 33/20.
2. Find the sums given below:
(i) 7 + 10 1/2 + 14 + … + 84
a=7, d=3.5, an=84. First find n: 84 = 7 + (n-1)3.5 => 77 = (n-1)3.5 => n-1=22 => n=23.
Sum S23 = 23/2 (7 + 84) = 23/2 (91) = 2093/2 = 1046.5.
(ii) 34 + 32 + 30 + … + 10
a=34, d=-2, an=10. First find n: 10 = 34 + (n-1)(-2) => -24 = (n-1)(-2) => n-1=12 => n=13.
Sum S13 = 13/2 (34 + 10) = 13/2 (44) = 13 × 22 = 286.
(iii) -5 + (-8) + (-11) + … + (-230)
a=-5, d=-3, an=-230. First find n: -230 = -5 + (n-1)(-3) => -225 = (n-1)(-3) => n-1=75 => n=76.
Sum S76 = 76/2 (-5 + (-230)) = 38(-235) = -8930.
3. In an AP:
(i) given a=5, d=3, an=50, find n and Sn.
50 = 5 + (n-1)3 => 45 = (n-1)3 => n-1=15 => n=16.
S16 = 16/2 (5 + 50) = 8(55) = 440.
(ii) given a=7, a13=35, find d and S13.
35 = 7 + (13-1)d => 28 = 12d => d=7/3.
S13 = 13/2 (7 + 35) = 13/2 (42) = 273.
(iii) given a12=37, d=3, find a and S12.
37 = a + (12-1)3 => 37 = a + 33 => a=4.
S12 = 12/2 (4 + 37) = 6(41) = 246.
(iv) given a3=15, S10=125, find d and a10.
a+2d=15. S10=10/2(2a+9d)=125 => 5(2a+9d)=125 => 2a+9d=25.
From first eq, a=15-2d. Sub into second: 2(15-2d)+9d=25 => 30-4d+9d=25 => 5d=-5 => d=-1.
a=15-2(-1)=17. a10 = a+9d = 17+9(-1) = 8.
(v) given d=5, S9=75, find a and a9.
S9=9/2(2a+8*5)=75 => 9/2(2a+40)=75 => 9(a+20)=75 => 3(a+20)=25 => 3a+60=25 => 3a=-35 => a=-35/3.
a9 = a+8d = -35/3 + 8(5) = -35/3 + 40 = 85/3.
(vi) given a=2, d=8, Sn=90, find n and an.
90 = n/2(2*2 + (n-1)8) => 180 = n(4+8n-8) = n(8n-4) => 8n²-4n-180=0 => 2n²-n-45=0.
(2n+9)(n-5)=0. Since n>0, n=5.
a5 = 2 + (5-1)8 = 2+32 = 34.
(vii) given a=8, an=62, Sn=210, find n and d.
210 = n/2(8+62) = n/2(70) = 35n => n=6.
a6=62 => 8+(6-1)d=62 => 5d=54 => d=54/5.
(viii) given an=4, d=2, Sn=-14, find n and a.
an=a+(n-1)2=4 => a=6-2n.
Sn=n/2(a+an) => -14=n/2(a+4).
Sub a: -14=n/2(6-2n+4) => -28=n(10-2n) => 2n²-10n-28=0 => n²-5n-14=0.
(n-7)(n+2)=0. Since n>0, n=7.
a=6-2(7) = -8.
(ix) given a=3, n=8, S=192, find d.
S8=192 = 8/2(2*3 + 7d) = 4(6+7d) => 48=6+7d => 42=7d => d=6.
(x) given l=28, S=144, and there are total 9 terms. Find a.
l=a9=28, n=9, S9=144.
S9=9/2(a+l) => 144=9/2(a+28) => 288=9(a+28) => 32=a+28 => a=4.
4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
a=9, d=8, Sn=636.
636 = n/2[2(9) + (n-1)8] = n/2[18 + 8n – 8] = n/2[10 + 8n] = n(5+4n) = 5n+4n².
4n² + 5n – 636 = 0.
Using quadratic formula, n = [-5 ± √(25 – 4(4)(-636))]/8 = [-5 ± √(25+10176)]/8 = [-5 ± √10201]/8 = [-5 ± 101]/8.
Since n must be positive, n = 96/8 = 12.
Answer: 12 terms.
… (Solutions for questions 5 to 20)
5. The 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.
a=5, l=45, Sn=400.
Sn = n/2(a+l) => 400 = n/2(5+45) = n/2(50) = 25n => n=16.
l=a16=45 => 5 + 15d = 45 => 15d = 40 => d=8/3.
6. The 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?
a=17, l=350, d=9.
l=an=350 => 17 + (n-1)9 = 350 => (n-1)9 = 333 => n-1=37 => n=38.
S38 = 38/2(17+350) = 19(367) = 6973.
7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
d=7, n=22, a22=149.
a22 = a + 21d => 149 = a + 21(7) => 149 = a + 147 => a=2.
S22 = 22/2(a + a22) = 11(2 + 149) = 11(151) = 1661.
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
a2=14, a3=18. So d = 18-14 = 4.
a = a2 – d = 14 – 4 = 10.
S51 = 51/2[2(10) + (51-1)4] = 51/2[20 + 200] = 51/2(220) = 51 × 110 = 5610.
9. If 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.
S7=49 => 7/2(2a+6d)=49 => 7(a+3d)=49 => a+3d=7.
S17=289 => 17/2(2a+16d)=289 => 17(a+8d)=289 => a+8d=17.
Subtracting the two equations: 5d=10 => d=2.
Substitute d: a+3(2)=7 => a=1.
Sn = n/2[2(1)+(n-1)2] = n/2[2+2n-2] = n/2(2n) = n².
10. Show that a₁, a₂, …, aₙ, … form an AP where aₙ is defined as below: (i) aₙ = 3 + 4n (ii) aₙ = 9 − 5n. Also find the sum of the first 15 terms in each case.
(i) aₙ = 3 + 4n
a₁=7, a₂=11, a₃=15. Difference is a₂-a₁=4. So it’s an AP with a=7, d=4.
S₁₅ = 15/2[2(7)+14(4)] = 15/2[14+56] = 15/2(70) = 525.
(ii) aₙ = 9 – 5n
a₁=4, a₂=-1, a₃=-6. Difference is a₂-a₁=-5. So it’s an AP with a=4, d=-5.
S₁₅ = 15/2[2(4)+14(-5)] = 15/2[8-70] = 15/2(-62) = -465.
11. If the sum of the first n terms of an AP is 4n − n², what is the first term? 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.
First term (a₁) = S₁ = 4(1) – 1² = 3.
Sum of first two terms (S₂) = 4(2) – 2² = 8 – 4 = 4.
Second term (a₂) = S₂ – S₁ = 4 – 3 = 1.
Third term (a₃) = S₃ – S₂ = (4*3-3²) – 4 = 3 – 4 = -1.
Tenth term (a₁₀) = S₁₀ – S₉ = (40-100) – (36-81) = -60 – (-45) = -15.
nth term (aₙ) = Sₙ – Sₙ₋₁ = (4n-n²) – [4(n-1)-(n-1)²] = 4n-n² – [4n-4-n²+2n-1] = 4n-n² – [6n-n²-5] = 5-2n.
12. Find the sum of the first 40 positive integers divisible by 6.
AP is 6, 12, 18, … a=6, d=6, n=40.
S₄₀ = 40/2[2(6)+39(6)] = 20[12+234] = 20(246) = 4920.
13. Find the sum of the first 15 multiples of 8.
AP is 8, 16, 24, … a=8, d=8, n=15.
S₁₅ = 15/2[2(8)+14(8)] = 15/2[16+112] = 15/2(128) = 15(64) = 960.
14. Find the sum of the odd numbers between 0 and 50.
AP is 1, 3, 5, …, 49. a=1, d=2, l=49.
49=1+(n-1)2 => 48=2(n-1) => n=25.
S₂₅ = 25/2(1+49) = 25/2(50) = 625.
15. A contract on construction job specifies a penalty for delay… Find how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
AP of penalties: 200, 250, 300, … a=200, d=50, n=30.
S₃₀ = 30/2[2(200)+29(50)] = 15[400+1450] = 15(1850) = 27750.
Answer: The contractor has to pay ₹27,750.
16. A sum of ₹700 is to be used to give seven cash prizes… Find the value of each of the prizes.
S₇=700, n=7, d=-20. Let the first prize be ‘a’.
700 = 7/2[2a+6(-20)] = 7/2[2a-120] = 7(a-60).
100 = a-60 => a=160.
Answer: The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.
17. In a school, students thought of planting trees… How many trees will be planted by the students?
Trees planted by Class I = 3 sections × 1 tree = 3.
Trees planted by Class II = 3 sections × 2 trees = 6.
AP is 3, 6, 9, …, to 12 terms (for Class XII). a=3, d=3, n=12.
S₁₂ = 12/2[2(3)+11(3)] = 6[6+33] = 6(39) = 234.
Answer: 234 trees will be planted.
18. A spiral is made up of successive semicircles… What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)
Length of semicircle = πr. Radii are 0.5, 1.0, 1.5, …
AP of lengths: 0.5π, 1.0π, 1.5π, …
a=0.5π, d=0.5π, n=13.
S₁₃ = 13/2[2(0.5π) + 12(0.5π)] = 13/2[π + 6π] = 13/2(7π).
= 13/2 × 7 × (22/7) = 13 × 11 = 143 cm.
19. 200 logs are stacked… How many rows are the 200 logs placed and how many logs are in the top row?
AP is 20, 19, 18, … a=20, d=-1, Sₙ=200.
200 = n/2[2(20)+(n-1)(-1)] = n/2[40-n+1] = n/2(41-n).
400 = 41n – n² => n² – 41n + 400 = 0 => (n-16)(n-25)=0.
If n=25, a₂₅ = 20+24(-1) = -4, which is not possible.
So, n=16 rows.
Logs in top row (a₁₆) = 20+15(-1) = 5.
20. In a potato race… What is the total distance the competitor has to run?
Distances from bucket: 5, 8, 11, … for 10 potatoes.
This forms an AP with a=5, d=3, n=10.
Competitor runs to the potato and back. Total distance = 2 × (Sum of distances).
S₁₀ = 10/2[2(5)+9(3)] = 5[10+27] = 5(37) = 185 m.
Total distance run = 2 × 185 = 370 m.
