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CBSE · NCERT

Class 10 Maths – Chapter 5: Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence where each term differs from the previous by a constant called the common difference (d). Key formulas: nth term aₙ = a + (n'1)d, Sum Sₙ = n/2[2a + (n'1)d] = n/2[a + l].

Exercises: 5.1–5.4·Total Questions: 49

Exercise 5.1 4 Questions – Identify APs

Q 1Identify AP

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) Taxi fare for each km: ₹15 for 1st km + ₹8 for each additional km. Sequence: 15, 23, 31, 39... d=8 ' AP

(ii) Amount of air in cylinder when vacuum pump removes ¼ each time. Not AP (geometric, ratio = 3/4).

Q 2–3Find a and d

Write first four terms of AP when: a=10, d=10 ' 10,20,30,40. a='2, d=0 ' '2,'2,'2,'2.

Q3: Check which are APs: (i) 2,4,8,16 ' not AP (geometric). (ii) 2,5/2,3,7/2 ' d=1/2 ' AP. (iii) '1.2,'3.2,'5.2,'7.2 ' d='2 ' AP.

Exercise 5.2 20 Questions – nth term (aₙ = a + (n'1)d)

Q 1–7Find Terms

Q1: Find nth term of 2,4,6,8... ' a=2, d=2. aₙ=2+2(n'1)=2n, a₁₅=30

Q2: Which term of 3,6,9,12... is 99? 3n=99 ' n=33

Q3: Which term of 15,12,9,6... is '27? a=15, d='3. 15+(n'1)('3)='27 ' n=15

Q4: Which term of 4,9,14,19... is 109? ' a=4, d=5. 4+5(n'1)=109 ' n=22

Q5: Find missing terms in AP: 2, __, 26 ' a=2, a₃=26. 2+2d=26 ' d=12. Missing:14

Q6: How many three-digit numbers are divisible by 7? First=105, last=994. 105+(n'1)7=994 ' n=128

Q7: Find 11th term from last of 10,7,4... '62. Last is '62. Reverse AP: a='62, d=3. a₁₁='62+10(3)='32

Q 8–20Word Problems

Q11: Which term of AP 3, 15, 27... is 132 more than its 54th term? a₅₄=3+53—12=639. aₙ=639+132=771. n=((771'3)/12)+1=65

Q16: 7th term is 4 times 2nd term. 12th term is 2 more than 3 times the 4th term. Find AP. ' a=2, d=3. AP: 2,5,8,11...

Q17: 17th term exceeds 10th term by 7. Find common difference. ' a₁₇'a₁₀=7d=7 ' d=1

Q20: Ramkali saved ₹5 on first day, ₹7 on second, ₹9 on third... How many days to save ₹715? a=5, d=2. Sₙ=n/2[10+2(n'1)]=715 ' n²+4n'715=0 ' 25 days

Exercise 5.3 20 Questions – Sum of n Terms (Sₙ)

Q 1–6Find Sum

Q1: Find sum of 2,7,12... to 10 terms. a=2, d=5. S₁₀=5[4+9—5]=5—49=245

Q2: Sum of '37,'33,'29... to 12 terms. S₁₂=6['74+11—4]=6—('30)='180

Q3: Given aₙ=3n+2, find sum of first 10 terms. a=5, a₁₀=32. S₁₀=5(5+32)=185

Q6: Sum of first 20 terms of AP with a=4, d=5. S₂₀=10[8+95]=1030

Q 7–20Word Problems

Q8: Find sum of first 51 terms where 2nd=14 and 3rd=18. d=4, a=10. S₅₁=5610

Q9: Sum of first 7 terms=49, sum of first 17 terms=289. Find sum of first n terms. ' a=3, d=2. Sₙ=n(n+2)

Q11: Sum of first n terms of AP is 4n'n². Find a, d, and Sₙ formula. ' Compare: Sₙ=n/2[2a+(n'1)d] with n(4'n). a=3, d='2.

Q18: Spiral made of semicircles with radii 0.5, 1.0, 1.5, 2.0 cm... Total length of 13 consecutive semicircles. ' π/2(0.5+1.0+1.5+...+6.5) = π/2 — S₁₃. AP: 0.5,1,1.5,... d=0.5. S₁₃=13/2[1+12—0.5]=13—3.5=45.5—π/2 cm

Q20: 200 logs stacked: 20 bottom, 19, 18... How many rows? Sₙ=n/2[40'(n'1)]=200 ' 20n'n(n'1)/2=200 ' n²'41n+400=0 ' n=16 or 25

Exercise 5.4 (Optional) 5 Questions

Q 1Advanced

Which term of AP 121, 117, 113... is its first negative term? a=121, d='4. aₙ<0 ' 121'4(n'1)<0 ' n>31.25. n=32. 32nd term = '3

Q 3Application

A ladder has rungs 25 cm apart. Top rung=45 cm, bottom=25 cm. Wood required for 2.5 m ladder? ' Number of rungs=250/25+1=11. Sum formula for AP: a=25, l=45, n=11. S₁₁=11/2(25+45)=385 cm

" Key Formulas: aₙ = a + (n'1)d · Sₙ = n/2[2a + (n'1)d] = n/2(a + l)

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