Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Idea | Formula |
|---|---|
| AP nth term | tₙ = a + (n − 1)d |
| AP recursive | t₁ = a, tₙ = tₙ₋₁ + d |
| Sum 1 + … + n | n(n + 1)/2 |
| GP nth term | tₙ = a rⁿ⁻¹ |
| GP recursive | t₁ = a, tₙ = r tₙ₋₁ |
| Triangular | n(n + 1)/2 |
| Fibonacci–Virahānka | Vₙ = Vₙ₋₁ + Vₙ₋₂ |
Try each problem first, then open the solution to check your working.
Write the first four terms of tₙ = 4n − 3 and the 25th term.
t₁ = 1, t₂ = 5, t₃ = 9, t₄ = 13.
t₂₅ = 100 − 3 = 97.
Answer: 1, 5, 9, 13; 97
Is 200 a term of tₙ = 3n + 2? Is 302?
3n + 2 = 200 gives n = 66, so 200 is the 66th term.
3n + 2 = 302 gives n = 100, so 302 is the 100th term.
Answer: both are terms
t₁ = 2, tₙ = 3tₙ₋₁ − 2. Find the first five terms.
t₂ = 4, t₃ = 10, t₄ = 28, t₅ = 82.
Answer: 2, 4, 10, 28, 82
V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂. Find V₉.
1, 2, 3, 5, 8, 13, 21, 34, 55.
Answer: 55
The AP is 7, 11, 15, … Find the 20th term and the term equal to 99.
a = 7, d = 4: t₂₀ = 7 + 19 × 4 = 83.
7 + 4(n − 1) = 99 gives n = 24.
Answer: 83; 24th term
In an AP, t₅ = 19 and t₉ = 35. Find a, d and t₂₀.
4d = 16, d = 4; a = 19 − 16 = 3.
t₂₀ = 3 + 19 × 4 = 79.
Answer: a = 3, d = 4, t₂₀ = 79
Find 1 + 2 + … + 60 and 41 + 42 + … + 60.
S₆₀ = 60 × 61/2 = 1830.
41 + … + 60 = 1830 − S₄₀ = 1830 − 820 = 1010.
Answer: 1830; 1010
How many multiples of 6 lie between 50 and 200?
54, 60, …, 198: (198 − 54)/6 + 1 = 25.
Answer: 25
A salary of ₹4,00,000 rises by ₹15,000 each year. After how many increments does it reach ₹6,25,000?
400000 + 15000k = 625000 gives k = 15.
Answer: 15 increments
Find the 8th term of 2, 6, 18, … and the term equal to 1458.
a = 2, r = 3: t₈ = 2 × 3⁷ = 4374.
2 × 3ⁿ⁻¹ = 1458 gives 3ⁿ⁻¹ = 729 = 3⁶, so n = 7.
Answer: 4374; 7th term
Is 81, −27, 9, −3, … a GP? Find tₙ.
Ratios: −27/81 = −1/3, 9/(−27) = −1/3, −3/9 = −1/3.
tₙ = 81(−1/3)ⁿ⁻¹.
Answer: yes, r = −1/3
A culture of 50 bacteria doubles every hour. How many after 6 hours? After n hours?
50 × 2⁶ = 3200; after n hours 50 × 2ⁿ.
Answer: 3200; 50 × 2ⁿ
A ball dropped from 10 m rebounds to 80% of its height each time. Find the height after the 3rd bounce.
10 × 0.8³ = 5.12.
Answer: 5.12 m
Stage 0 is a unit square; each stage removes the middle ninth of every square. Find the area left at stage 3.
Area × 8/9 each stage: (8/9)³ = 512/729.
Answer: 512/729 ≈ 0.70
Which of 5, 10, 15, 20, … ; 5, 10, 20, 40, … ; 1, 4, 9, 16, … are AP, GP, or neither?
5, 10, 15, 20: AP (d = 5).
5, 10, 20, 40: GP (r = 2).
1, 4, 9, 16: neither (differences 3, 5, 7; ratios 4, 2.25, …).
Answer: AP, GP, neither
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Natural: add 1 (7, 8, 9). Odd: add 2 (13, 15, 17). Triangular: add 2, 3, 4, 5, 6, … (28, 36, 45). Square: add 3, 5, 7, 9, 11, … (49, 64, 81).
15, 21, 28, 36.
Next four terms: 16, 19, 22, 25 (tₙ = 3n − 2). Running sums: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.
u₅₃ = 105, u₁₀₈ = 215, u₁₁₇₀ = 2339.
3, 8, 13, 18, 23, 28; s₁₀₀ = 498; s₁₀₀₀ = 4998. 308: 5n = 310, n = 62: the 62nd term. 473: 5n = 475, n = 95: the 95th term. The text works with 471: 5n = 473 gives n = 94.6, not a natural number, so 471 is not a term. (The printed 473 appears to be a slip for 471.) n must be a natural number because a term has a whole-number position.
tₙ = n².
(i) −4, −1, 2, 29, 47, 143. (ii) 3n − 7 = 332: n = 113, the 113th term. (iii) 3n − 7 = 557 gives n = 188: yes, the 188th term.
Terms: 1, 5, 13, 29, 61, 125, 253, … 133 is skipped (125 then 253). Also uₙ + 3 = 2ⁿ⁺¹ and 136 is not a power of 2. 133 is not a term.
55, 89.
Natural numbers (d = 1), odd numbers (d = 2), 1, 4, 7, 10, … (d = 3), −7, −3, 1, 5, … (d = 4) are APs. Triangular (differences 2, 3, 4, …), squares (3, 5, 7, …), primes and 1, 1/2, 1/3, … are not (the differences are not constant).
(i) d = 3: tₙ = 2 + 3(n − 1) = 3n − 1. (ii) d = 4: tₙ = −5 + 4(n − 1) = 4n − 9. The points (n, tₙ) lie on a straight line.
(i) a = 1/2, d = 2: tₙ = 1/2 + 2(n − 1) = 2n − 3/2. (ii) a = 1.5, d = 2: tₙ = 2n − 0.5.
2, 5, 8, …: t₁ = 2, tₙ = tₙ₋₁ + 3. −5, −1, 3, …: t₁ = −5, tₙ = tₙ₋₁ + 4. 1/2, 5/2, …: t₁ = 1/2, tₙ = tₙ₋₁ + 2. 1.5, 3.5, …: t₁ = 1.5, tₙ = tₙ₋₁ + 2.
tₙ = 4n − 3: stage 5: 17; 6: 21; 10: 37; 11: 41; 12: 45; 20: 77; stage n: 4n − 3.
tₙ = 200 + 40n; for n = 10: ₹600.
Yes: 2S = 100 × 101, so S = 5050.
S₂₀ = 210, S₅₀ = 1275, S₁₀₀₀ = 500500.
t₁₀ = 55, t₁₇ = 153, t₈₀ = 3240.
(i) −1, 2, 5, 8, 11. (ii) −3, −8, −13, −18, −23. (iii) 2, 3, 6, 11, 18.
t₁₀ = 47; t₁₅ = 72.
5n − 3 = 97: n = 20 (yes, 20th term). 5n − 3 = 172: n = 35 (yes, 35th term).
5n = 610, n = 122.
−5, −2, 1, 4, 7 (tₙ = 3n − 8). 3n − 8 = 52 gives n = 20: yes, the 20th term.
T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81.
a = 3, d = 5: t₁₀ = 3 + 45 = 48; t₂₆ = 3 + 125 = 128.
d = −3: tₙ = 24 − 3n. 24 − 3n = −81 gives n = 35. 24 − 3n = 0 gives n = 8, a natural number, so 0 is the 8th term.
tₙ = 11 − 3(n − 1) = 14 − 3n. Recursive: t₁ = 11, tₙ = tₙ₋₁ − 3.
a + 2d = 12 and a + 49d = 106: 47d = 94, d = 2, a = 8. t₂₉ = 8 + 28 × 2 = 64.
12, 15, …, 99: n = (99 − 12)/3 + 1 = 30. Sum = 30 × (12 + 99)/2 = 1665.
5,00,000 + 20,000k = 7,00,000 gives k = 10. So after 10 increments, that is, in the 11th year the salary is ₹7,00,000.
S₂₅ = 25 × 26/2 = 325 marbles.
tₙ = 3 × 2ⁿ⁻¹: stage 5: 48; 6: 96; 10: 1536; 11: 3072; 12: 6144; 20: 1,572,864; stage n: 3 × 2ⁿ⁻¹. Here each stage doubles (multiply by 2, a GP); in Fig. 8.3 a fixed 4 is added (an AP).
Yes in each case: ratios 2; 3; −1 respectively.
(i) r = 5: tₙ = 2 × 5ⁿ⁻¹. (ii) r = 2/3: tₙ = 4(2/3)ⁿ⁻¹. (iii) r = −1/2: tₙ = 3(−1/2)ⁿ⁻¹.
t₁ = 3, tₙ = 10 tₙ₋₁.
(a) 1, 3, 9, 27. (b) 81 and 243. (c) tₙ = 3ⁿ, since each black triangle is replaced by three. (d) Area: 1, 3/4, 9/16, 27/64; stage 4: 81/256; stage 5: 243/1024; stage n: (3/4)ⁿ, which gets closer and closer to 0.
Heights: 18, 13.5, 10.125, 7.594, 5.695. Sixth: 4.271; seventh: 3.203 < 4. So 7 bounces.
t₁₂ = t₈ × 2⁴ = 192 × 16 = 3072.
a = 5, r = 5: tₙ = 5ⁿ; t₁₀ = 5¹⁰ = 9,765,625.
2, 4, 10, 28, 82, 244, 730. Also tₙ − 1 = 3ⁿ⁻¹ (tₙ = 3ⁿ⁻¹ + 1), and 3ⁿ⁻¹ = 729 gives n = 7. It is the 7th term.
tₙ = 2 × 3ⁿ⁻¹ = 4374: 3ⁿ⁻¹ = 2187 = 3⁷, n = 8. Recursive: t₁ = 2, tₙ = 3tₙ₋₁.
(i) 80 × 0.6⁵ = 80 × 0.07776 = 6.2208 m. (ii) The first fall is 80 m; each of bounces 1 to 5 adds an up-and-down of 2 × 80 × 0.6ᵏ: total = 80 + 160(0.6 + 0.36 + 0.216 + 0.1296 + 0.07776) = 80 + 160 × 1.38336 = 301.34 m.
r = √2: 2(√2)ⁿ⁻¹ = 128 gives (√2)ⁿ⁻¹ = 64 = (√2)¹², n − 1 = 12, n = 13.
1, 8, 64, 512.
4096 and 32768.
tₙ = 8ⁿ (n = stage number from 0). Recursive: t₀ = 1, tₙ = 8tₙ₋₁.
Each stage keeps 8 of 9 equal squares: 8/9, 64/81, 512/729; stage 4: 4096/6561; stage 5: 32768/59049. Explicit: (8/9)ⁿ; recursive: s₀ = 1, sₙ = (8/9)sₙ₋₁. As n increases the area tends to 0.
5d = 35, d = 7; a = 38 − 70 = −32. t₃₁ = −32 + 30 × 7 = 178.
2d = 12, d = 6; a + 12 = 16, a = 4. The AP is 4, 10, 16, 22, ….
105, 112, …, 994: n = (994 − 105)/7 + 1 = 128.
12, 16, …, 248: n = (248 − 12)/4 + 1 = 60.
t₅ = 4t₃ gives r² = 4, r = ±2. r = 2: a(1 + 2) = −4, a = −4/3: −4/3, −8/3, −16/3, …. r = −2: a(1 − 2) = −4, a = 4: 4, −8, 16, −32, ….
k terms starting at m: k(2m + k − 1)/2 = 100, so k(2m + k − 1) = 200 where the two factors have opposite parity. k = 5: 2m + 4 = 40, m = 18: 18 + 19 + 20 + 21 + 22 = 100. k = 8: 2m + 7 = 25, m = 9: 9 + 10 + … + 16 = 100. (k = 1 gives 100 itself.) So there are two ways with at least two terms.
30 × 2ⁿ: 120, 480, 30 × 2ⁿ.
a + 5d = 12 and a + 7d = 22: 2d = 10, d = 5, a = −13. The terms: −13, −8, −3.
n(n + 1)/2 > 1000. S₄₄ = 990 and S₄₅ = 1035. So n = 45.
r = 4: 2 × 4ⁿ⁻¹ = 2¹⁷ gives 4ⁿ⁻¹ = 2¹⁶ = 4⁸, n = 9. Explicit tₙ = 2 × 4ⁿ⁻¹ = 2²ⁿ⁻¹; recursive t₁ = 2, tₙ = 4tₙ₋₁.
Write the terms a/r, a, ar. Product a³ = −1, so a = −1. Sum: −(1/r + 1 + r) = 13/12, which gives 12r² + 25r + 12 = 0, r = −3/4 or −4/3. For r = −3/4: terms 4/3, −1, 3/4. For r = −4/3: terms 3/4, −1, 4/3.
x = ar³, y = ar⁹, z = ar¹⁵. y/x = r⁶ and z/y = r⁶, so the ratios are equal: x, y, z form a GP with ratio r⁶.
a(1 + r + r²) = 26 and a²(1 + r² + r⁴) = 364. Since 1 + r² + r⁴ = (1 + r + r²)(1 − r + r²), dividing gives a(1 − r + r²) = 14. Then (1 + r + r²)/(1 − r + r²) = 13/7, so 3r² − 10r + 3 = 0, r = 3 or 1/3. The terms are 2, 6, 18 or 18, 6, 2.
P₃ = 4, P₄ = 8, P₅ = 16, P₆ = 32, P₇ = 64, P₈ = 128. Since Pₙ = Pₙ₋₁ + (P₁ + … + Pₙ₋₂ + 1) = 2Pₙ₋₁ for n ≥ 3: Pₙ = 2Pₙ₋₁, explicit Pₙ = 2ⁿ⁻¹.
W₃ = 3, W₄ = 5, W₅ = 8, W₆ = 13, W₇ = 21, W₈ = 34. This is the Virahānka–Fibonacci sequence.