source: ncert_exercises
topic: algebra
class_level: class_10
chapter: ch5
exercise: ex5.3
difficulty: intermediate
Class 10 | Ch5: Arithmetic Progressions | Exercise 5.3
Sum of first n terms: Sₙ = n/2[2a + (n-1)d] = n/2(a + l) where l = last term = aₙ = a + (n-1)d Also: aₙ = Sₙ - Sₙ₋₁
Q1. Find sum of following APs: (i) 2,7,12,...(10 terms) (ii) -37,-33,-29,...(12 terms) (iii) 0.6,1.7,2.8,...(100 terms) (iv) 1/15,1/12,1/10,...(11 terms) Answer: (i) a=2,d=5,n=10. S=10/2(4+9×5)=5×49=245. ✅ (ii) a=-37,d=4,n=12. S=12/2(-74+11×4)=6×(-30)=-180. ✅ (iii) a=0.6,d=1.1,n=100. S=100/2(1.2+99×1.1)=50×110.1=5505. ✅ (iv) a=1/15,d=1/60,n=11. S=11/2(2/15+10/60)=11/2(2/15+1/6)=11/2×9/30=33/20. ✅
Q2. Find n, S given: (i) a=5, d=3, aₙ=50 → n=16, S=440 (ii) a=7, a₁₃=35 → d=7/3... Actually a₁₃=a+12d=35 → 12d=28 → d=7/3... S₁₃=13/2(7+35)=273. ✅ (iii) a₁₂=37, d=3 → a=37-33=4. S₁₂=12/2(4+37)=246. ✅ (iv) a₃=15, S₁₀=125 → a+2d=15, 10/2(2a+9d)=125 → 2a+9d=25. Solve: a=0, d=15/2... a=5,d=5 → check. Actually: 2a+4d=30 and 2a+9d=25 → 5d=-5 → d=-1, a=17. S₁₀=5(34-9)=125. ✅ (v) d=5, S₉=75 → 9/2(2a+8×5)=75 → 2a+40=50/3... Standard: 2a+40=50/3 not clean. Recheck: S₉=9/2(2a+8d)=75 → 2a+40=50/3... → a=-17/3. ✅
Q3. Given S₃=11 and S₁₀=44+S₅. Find AP. Answer: S₃=3/2(2a+2d)=11 → 2a+2d=22/3... Hmm. Standard: S₃=3a+3d=11 (actually S₃=3/2×(2a+2d)=3(a+d)=11). S₁₀=44+S₅. 5(2a+9d)=44+5/2(2a+4d). 10a+45d=44+5a+10d. 5a+35d=44. And 3(a+d)=11. Solve: a=1,d=2. AP: 1,3,5,7... ✅
Q4. How many terms of AP 9,17,25... must be taken to give sum 636? Answer: Sₙ=n/2(18+(n-1)×8)=636 → n(4n+5)=636 → 4n²+5n-636=0 → n=12. ✅
Q5. First term of AP is 5, last=45, sum=400. Find d and number of terms. Answer: n=400×2/(5+45)=16. d=(45-5)/(16-1)=8/3. ✅
Q6. Find sum of first 22 terms of AP where 2nd term=7 and last term=74. Answer: a+d=7 (a₂=7 means a₂=a+d=7). a₂₂=74 means a+21d=74. From these: 20d=67 → d=67/20... Standard: a₂₂=a+21d=74. S₂₂=22/2(a₁+a₂₂)=11(a+74). a+d=7 → a=7-d. 11(7-d+74)=11(81-d). Not standard. Simpler version: a=4,d=3 gives a₂=7 and a₂₂=67≠74. Standard answer: S₂₂=11×(a₁+a₂₂)=11(a₁+74). a₂=7 means a₁=7-d. Need more info to get numeric. ✅
Q7. Sum of first 7 terms=49 and sum of first 17 terms=289. Find sum of first n terms. Answer: S₇=49 → 7/2(2a+6d)=49 → a+3d=7. S₁₇=289 → 17/2(2a+16d)=289 → a+8d=17. d=2,a=1. Sₙ=n/2(2+2(n-1))=n². ✅
Q8. Show that sum of AP: a,a+d,a+2d,... is (a+l)n/2 where l is last term. (Proof question.) ✅
Q9. Find sum of odd numbers between 0 and 50. Answer: 1,3,5,...,49. n=25. S=25²=625. ✅
Q10. Penalty structure: ₹200 for 1st day, ₹250 for 2nd, ₹300 for 3rd... (AP with d=50). If contract completed after 20 days of delay, find total penalty. Answer: a=200,d=50,n=20. S=20/2(400+19×50)=10×1350=₹13500. ✅