NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 5.1
In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) Taxi fare for each km when fare is ₹15 for first km and ₹8 for each subsequent km: 15, 23, 31, 39... Common difference = 8 → Yes, AP
(ii) Amount of air in a cylinder when a vacuum pump removes 1/4 of the air remaining each time: Not AP because the amount removed decreases each time (it's a GP) → No, not AP
(iii) Cost of digging for each metre when cost for first metre is ₹150 and rises by ₹50 for each subsequent metre: 150, 200, 250, 300... d = 50 → Yes, AP
(iv) Amount in account after each year at 8% SI on ₹10,000: Interest = 800/yr. Amount: 10800, 11600, 12400... d = 800 → Yes, AP
Answer
Write first four terms of the AP when the first term a and the common difference d are given:
(i) a = 10, d = 10: 10, 20, 30, 40
(ii) a = –2, d = 0: –2, –2, –2, –2
(iii) a = 4, d = –3: 4, 1, –2, –5
(iv) a = –1, d = 1/2: –1, –1/2, 0, 1/2
Answer
For the following APs, write the first term and the common difference:
(i) 3, 1, –1, –3...: a = 3, d = –2
(ii) –5, –1, 3, 7...: a = –5, d = 4
(iii) 1/3, 5/3, 9/3, 13/3...: a = 1/3, d = 4/3
(iv) 0.6, 1.7, 2.8, 3.9...: a = 0.6, d = 1.1
Answer
Which 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...: 4–2=2, 8–4=4 (not constant) → Not AP
(ii) 2, 5/2, 3, 7/2...: d = 1/2 → AP. Next: 4, 9/2, 5
(iii) –1.2, –3.2, –5.2, –7.2...: d = –2 → AP. Next: –9.2, –11.2, –13.2
(iv) –10, –6, –2, 2...: d = 4 → AP. Next: 6, 10, 14
(v) 3, 3+√2, 3+2√2, 3+3√2...: d = √2 → AP. Next: 3+4√2, 3+5√2, 3+6√2
(vi) 0.2, 0.22, 0.222...: 0.22–0.2=0.02, 0.222–0.22=0.002 (not constant) → Not AP
Answer
(vii) 0, –4, –8, –12...: d = –4
⇒ AP. Next: –16, –20, –24
(viii) –1/2, –1/2, –1/2...: d = 0
⇒ AP. Next: –1/2, –1/2, –1/2
Exercise 5.2
Q1. Fill in the blanks
(i) a = 7, d = 3, n = 8, aₙ = a + (n–1)d = 7 + 7×3 = 28
(ii) a = –18, n = 10, aₙ = 0: 0 = –18 + 9d → d = 2
(iii) a = ?, d = –3, n = 18, aₙ = –5: –5 = a + 17(–3) → a = –5 + 51 = 46
(iv) a = –18.9, d = 2.5, aₙ = 3.6, n = ?: 3.6 = –18.9 + (n–1)(2.5) → n–1 = 22.5/2.5 = 9 → n = 10
(v) a = 3.5, d = 0, n = 105, aₙ = 3.5 + 0 = 3.5
Q2. Choose the correct choice
(i) 30th term of AP 10, 7, 4...: a₃₀ = 10 + 29(–3) = 10 – 87 = –77 → Answer (C)
(ii) 11th term of AP –3, –1/2, 2, 9/2...: d = 5/2. a₁₁ = –3 + 10(5/2) = –3 + 25 = 22 → Answer (B)
Q3. In the following APs, find the missing terms
(i) 2, __, 26: d = (26–2)/2 = 12. Missing term = 14
(ii) __, 13, __, 3: d = (3–13)/2 = –5. Terms: 18, 13, 8, 3
(iii) 5, __, __, 9½: d = (9.5–5)/3 = 1.5. Terms: 5, 6.5, 8, 9.5
(iv) –4, __, __, __, __, 6: d = (6–(–4))/5 = 2. Terms: –4, –2, 0, 2, 4, 6
(v) __, 38, __, __, __, –22: d = (–22–38)/4 = –15. Terms: 53, 38, 23, 8, –7, –22
Which term of the AP 3, 8, 13, 18... is 78?
Answer
aₙ = 3 + (n–1)5 = 78
⇒ 5n – 2 = 78
⇒ n = 16. 16th term
Q5. Find the number of terms in each of the following APs
(i) 7, 13, 19..., 205: d = 6. 205 = 7 + (n–1)6 → n = 34
(ii) 18, 15½, 13..., –47: d = –5/2. –47 = 18 + (n–1)(–5/2) → n = 27
Check whether –150 is a term of the AP 11, 8, 5, 2...
Answer
–150 = 11 + (n–1)(–3)
⇒ –161 = –3(n–1)
⇒ n–1 = 161/3 = 53.67. Not a whole number
⇒ –150 is not a term.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Answer
a₁₆ – a₁₁ = 5d
⇒ 73 – 38 = 5d
⇒ d = 7. a₁₁ = a + 10d
⇒ 38 = a + 70
⇒ a = –32.
a₃₁ = –32 + 30(7) = 178
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer
a₃ = a + 2d = 12, a₅₀ = a + 49d = 106. Subtracting: 47d = 94
⇒ d = 2. a = 8.
a₂₉ = 8 + 28(2) = 64
Exercise 5.3
Q1. Find the sum of the following APs
(i) 2, 7, 12..., to 10 terms: S = n/2[2a + (n–1)d] = 10/2[4 + 45] = 5 × 49 = 245 (wait: 2a=4, (n-1)d=9×5=45, sum=10/2 × 49 = 245. Actually let me recalculate: S₁₀ = 10/2[2(2) + 9(5)] = 5[4+45] = 5×49 = 245)
(ii) –37, –33, –29..., to 12 terms: S = 12/2[2(–37) + 11(4)] = 6[–74 + 44] = 6(–30) = –180
(iii) 0.6, 1.7, 2.8..., to 100 terms: d = 1.1. S = 100/2[2(0.6) + 99(1.1)] = 50[1.2 + 108.9] = 50 × 110.1 = 5505
(iv) 1/15, 1/12, 1/10..., to 11 terms: d = 1/12 – 1/15 = 1/60. S = 11/2[2/15 + 10/60] = 11/2[2/15 + 1/6] = 11/2[4/30 + 5/30] = 11/2 × 9/30 = 11/2 × 3/10 = 33/20
Q2. Find the sums
(i) 7 + 10½ + 14 +... + 84: d = 3.5. 84 = 7 + (n–1)(3.5) → n = 23. S = 23/2(7+84) = 23/2 × 91 = 1046.5
(ii) 34 + 32 + 30 +... + 10: d = –2. 10 = 34 + (n–1)(–2) → n = 13. S = 13/2(34+10) = 13/2 × 44 = 286
(iii) –5 + (–8) + (–11) +... + (–230): d = –3. –230 = –5 + (n–1)(–3) → n = 76. S = 76/2(–5 + (–230)) = 38 × (–235) = –8930
In an AP: (i) a = 5, d = 3, aₙ = 50, find n and Sₙ.
Answer
50 = 5 + (n–1)3
⇒ n = 16. S₁₆ = 16/2(5+50) = 8 × 55 = 440
(ii) a = 7, a₁₃ = 35, find d and S₁₃: 35 = 7 + 12d
⇒ d = 7/3. S₁₃ = 13/2(7+35) = 273
(iii) a₁₂ = 37, d = 3, find a and S₁₂: 37 = a + 11(3)
⇒ a = 4. S₁₂ = 12/2(4+37) = 246
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer
a₂₂ = a + 21(7) = 149
⇒ a = 2. S₂₂ = 22/2(2 + 149) = 11 × 151 = 1661
Show that a₁, a₂..., aₙ form an AP where aₙ = 3 + 4n. Also find the sum of the first 15 terms.
Answer
aₙ = 3 + 4n. a₁ = 7, a₂ = 11, a₃ = 15. d = 4 (constant)
⇒ AP ✓
S₁₅ = 15/2(a₁ + a₁₅) = 15/2(7 + 63) = 15/2 × 70 = 525
Important Extra Questions
The sum of the first n terms of an AP is given by Sₙ = 3n² + 5n. Find the AP and its 25th term.
Answer
S₁ = a₁ = 3 + 5 = 8. S₂ = 12 + 10 = 22
⇒ a₂ = 22 – 8 = 14. d = 14 – 8 = 6.
AP: 8, 14, 20, 26...
a₂₅ = 8 + 24(6) = 152
How many terms of the AP 24, 21, 18... must be taken so that their sum is 78?
Answer
Sₙ = n/2[2(24) + (n–1)(–3)] = n/2[48 – 3n + 3] = n/2[51 – 3n] = 78
n(51 – 3n) = 156
⇒ 3n² – 51n + 156 = 0
⇒ n² – 17n + 52 = 0
⇒ (n–4)(n–13) = 0
n = 4 or n = 13 (both valid — the AP has negative terms after the 9th, so partial sums can repeat)
Find the sum of all two-digit numbers divisible by 7.
Answer
AP: 14, 21, 28..., 98. a = 14, d = 7, l = 98. n = (98–14)/7 + 1 = 13.
S = 13/2(14 + 98) = 13/2 × 112 = 728