Chapter 8 exercise answers: Predicting What Comes Next: Exploring Sequences and Progressions

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(i) tn = 3n – 4

1. The rule tₙ = 3n − 4 means: take the position n, multiply by 3, then subtract 4.
2. n = 1: 3×1 − 4 = −1. n = 2: 3×2 − 4 = 2. n = 3: 3×3 − 4 = 5.
3. n = 4: 3×4 − 4 = 8. n = 5: 3×5 − 4 = 11.

Answer−1, 2, 5, 8, 11

(ii) tn = 2 – 5n

1. The rule tₙ = 2 − 5n means: take the position n, multiply by 5, then subtract from 2.
2. n = 1: 2 − 5 = −3. n = 2: 2 − 10 = −8. n = 3: 2 − 15 = −13.
3. n = 4: 2 − 20 = −18. n = 5: 2 − 25 = −23.

Answer−3, −8, −13, −18, −23

(iii) tn = n² – 2n + 3

1. The rule tₙ = n² − 2n + 3 needs n squared, then 2n subtracted, then 3 added.
2. n = 1: 1 − 2 + 3 = 2. n = 2: 4 − 4 + 3 = 3. n = 3: 9 − 6 + 3 = 6.
3. n = 4: 16 − 8 + 3 = 11. n = 5: 25 − 10 + 3 = 18.

Answer2, 3, 6, 11, 18

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1. Substitute n = 10 into tₙ = 5n − 3: 5×10 − 3 = 50 − 3 = 47.
2. Substitute n = 15: 5×15 − 3 = 75 − 3 = 72.

AnswerThe 10th term is 47 and the 15th term is 72.

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1. To test membership, set tₙ equal to the number and solve for n; n must come out a whole counting number.
2. For 97: 5n − 3 = 97, so 5n = 100, n = 20. This is a whole number, so 97 is the 20th term.
3. For 172: 5n − 3 = 172, so 5n = 175, n = 35. This is also a whole number, so 172 is the 35th term.

AnswerBoth 97 and 172 are terms of the sequence (the 20th and 35th terms).

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1. Set the rule equal to 607: 5n − 3 = 607.
2. Add 3 to both sides: 5n = 610.
3. Divide by 5: n = 122.

Answer607 is the 122nd term.

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1. Start at the seed t₁ = −5, and keep adding 3 to get each next term.
2. t₁ = −5, t₂ = −5+3 = −2, t₃ = −2+3 = 1, t₄ = 1+3 = 4, t₅ = 4+3 = 7.
3. This is really an arithmetic step, so the explicit rule is tₙ = −5 + (n−1)×3 = 3n − 8.
4. To check 52, solve 3n − 8 = 52: 3n = 60, so n = 20, a whole number.

AnswerThe first five terms are −5, −2, 1, 4, 7. Yes, 52 is a term — it is the 20th term.

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1. Each term (from the 4th onward) is the sum of the three terms just before it, so add three seeds to start.
2. T4 = T3+T2+T1 = 4+2+1 = 7.
3. T5 = T4+T3+T2 = 7+4+2 = 13.
4. T6 = T5+T4+T3 = 13+7+4 = 24.
5. T7 = T6+T5+T4 = 24+13+7 = 44.
6. T8 = T7+T6+T5 = 44+24+13 = 81.

AnswerT4 = 7, T5 = 13, T6 = 24, T7 = 44, T8 = 81.

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1. First term a = 3, common difference d = 8 − 3 = 5.
2. nth term formula: tₙ = a + (n−1)d = 3 + (n−1)×5.
3. 10th term: 3 + 9×5 = 3 + 45 = 48.
4. 26th term: 3 + 25×5 = 3 + 125 = 128.

AnswerThe 10th term is 48 and the 26th term is 128.

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1. First term a = 21, common difference d = 18 − 21 = −3, so tₙ = 21 + (n−1)(−3) = 24 − 3n.
2. For −81: 24 − 3n = −81, so 3n = 105, n = 35. A whole positive number, so −81 is the 35th term.
3. For 0: 24 − 3n = 0, so 3n = 24, n = 8. A whole positive number, so 0 is also a term — the 8th term.

Answer−81 is the 35th term of the AP. Yes, 0 is a term too — it is the 8th term.

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1. First term a = 11, common difference d = 8 − 11 = −3.
2. Explicit formula: tₙ = a + (n−1)d = 11 + (n−1)(−3) = 14 − 3n.
3. Recursive rule: t₁ = 11, and tₙ₊₁ = tₙ − 3 for n ≥ 1.

AnswerExplicit: tₙ = 14 − 3n. Recursive: t₁ = 11, tₙ₊₁ = tₙ − 3 for n ≥ 1.

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1. 3rd term: a + 2d = 12. Last (50th) term: a + 49d = 106.
2. Subtract the first equation from the second: 47d = 94, so d = 2.
3. Substitute back: a + 2(2) = 12, so a = 8.
4. 29th term: a + 28d = 8 + 28×2 = 8 + 56 = 64.

AnswerThe 29th term is 64.

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1. The 2-digit multiples of 3 form an AP: 12, 15, 18, ..., 99, with a = 12 and d = 3.
2. Number of terms: 99 = 12 + (n−1)×3, so (n−1) = 29, n = 30.
3. Sum = (number of terms)/2 × (first term + last term) = 30/2 × (12 + 99) = 15 × 111 = 1665.

AnswerThere are 30 such numbers, and their sum is 1665.

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1. The salary rises by a fixed amount each year, so it is an AP with a = ₹5,00,000 and d = ₹20,000.
2. The rise needed is ₹7,00,000 − ₹5,00,000 = ₹2,00,000.
3. Number of increments (years) needed = ₹2,00,000 ÷ ₹20,000 = 10.
4. So the income reaches ₹7,00,000 after 10 years, once 10 increments have been added to the starting salary.

AnswerHis income reached ₹7,00,000 after 10 years.

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1. The rows hold 1, 2, 3, ..., 25 marbles — the counting numbers up to 25.
2. Sum of the first n counting numbers is n(n+1)/2, so total = 25×26/2.
3. 25×26 = 650, and 650/2 = 325.

AnswerThe child uses 325 marbles in all.

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1. 8th term: a×r⁷ = 192, with r = 2, so a×128 = 192, giving a = 3/2.
2. 12th term is 4 positions past the 8th, so multiply by r four more times: 192 × 2⁴ = 192 × 16 = 3072.

AnswerThe 12th term is 3072.

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1. First term a = 5, common ratio r = 25/5 = 5.
2. nth term formula: tₙ = a×rⁿ⁻¹ = 5×5ⁿ⁻¹ = 5ⁿ.
3. 10th term: 5¹⁰ = 9765625.

AnswerThe 10th term is 9765625, and the nth term is 5ⁿ.

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1. Subtract 1 from every term: if uₙ = tₙ − 1, then uₙ₊₁ = tₙ₊₁ − 1 = (3tₙ−2) − 1 = 3(tₙ−1) = 3uₙ.
2. So uₙ is a GP with u₁ = t₁ − 1 = 1 and ratio 3, giving uₙ = 3ⁿ⁻¹, so tₙ = 3ⁿ⁻¹ + 1.
3. Set tₙ = 730: 3ⁿ⁻¹ + 1 = 730, so 3ⁿ⁻¹ = 729 = 3⁶, giving n − 1 = 6, n = 7.
4. Check by walking: 2, 4, 10, 28, 82, 244, 730 — the 7th term is indeed 730.

Answer730 is the 7th term.

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1. First term a = 2, common ratio r = 6/2 = 3, so the explicit formula is tₙ = 2×3ⁿ⁻¹.
2. Recursive formula: t₁ = 2, tₙ₊₁ = 3×tₙ for n ≥ 1.
3. Set 2×3ⁿ⁻¹ = 4374: 3ⁿ⁻¹ = 2187 = 3⁷, so n − 1 = 7, n = 8.

Answer4374 is the 8th term. Explicit: tₙ = 2×3ⁿ⁻¹. Recursive: t₁ = 2, tₙ₊₁ = 3tₙ for n ≥ 1.

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(i) What height does the ball reach after the 5th bounce?

1. Each bounce reaches 60% of the one before, so the bounce heights form a GP with first bounce height a = 80×0.6 = 48 and ratio r = 0.6.
2. 5th bounce height = a×r⁴ = 48×(3/5)⁴ = 48×81/625 = 3888/625.

Answer3888/625 m, which is 6.2208 m

(ii) total vertical distance … hits the ground for the 6th time?

1. The ball first touches the ground after the drop (1st landing). Landings 2 through 6 each follow a bounce up and a fall back down.
2. So the 6th landing uses the up-and-down travel of bounces 1 to 5: heights 48, 28.8, 17.28, 10.368, 6.2208.
3. Total distance = the drop + twice the sum of these 5 bounce heights = 80 + 2×(48+28.8+17.28+10.368+6.2208).
4. The sum of the 5 heights is 110.6688, so total = 80 + 221.3376 = 301.3376 m = 188336/625 m.

Answer188336/625 m, which is 301.3376 m

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1. Check it is a GP: 2√2 ÷ 2 = √2, and 4 ÷ 2√2 = √2. So the first term a = 2 and the common ratio r = √2.
2. The nth term is a·rⁿ⁻¹, so 2·(√2)ⁿ⁻¹ = 128.
3. Divide both sides by 2: (√2)ⁿ⁻¹ = 64.
4. Write both sides as powers of 2: √2 = 2^1/2 and 64 = 2⁶, so 2^(n−1)/2 = 2⁶.
5. Match the exponents: (n − 1)/2 = 6, so n − 1 = 12 and n = 13.

Answer128 is the 13th term.

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(i) How many red squares are there in Stages 0 to 3?

1. Stage 0 is one whole red square: 1.
2. Each stage keeps 8 of the 9 small squares made from every red square of the stage before, so multiply by 8 each time.
3. Stage 1 = 1 × 8 = 8. Stage 2 = 8 × 8 = 64. Stage 3 = 64 × 8 = 512.

AnswerStage 0: 1, Stage 1: 8, Stage 2: 64, Stage 3: 512.

(ii) predict the number of red squares in Stages 4 and 5?

1. Keep multiplying by 8: Stage 4 = 512 × 8 = 4096.
2. Stage 5 = 4096 × 8 = 32768.

AnswerStage 4: 4096 squares, Stage 5: 32768 squares.

(iii) a rule for the number of red squares at the nth stage?

1. The count of red squares is a GP with first term 1 (at stage 0) and common ratio 8.
2. Explicit formula: number of red squares at stage n = 8ⁿ.
3. Recursive formula: (number at stage n) = 8 × (number at stage n − 1), for n ≥ 1, with stage 0 = 1.

AnswerExplicit: 8ⁿ. Recursive: a_n = 8·a_n−1 (n ≥ 1), a₀ = 1; common ratio 8.

(iv) area of the red region in Stages 1, 2 and 3?

1. Each side is cut into 3 equal parts, so each small square's area is (1/3)² = 1/9 of the square it came from, and 8 of these 9 pieces are kept, so 8/9 of the area survives each stage.
2. Stage 1 area = 1 × 8/9 = 8/9. Stage 2 = 8/9 × 8/9 = 64/81. Stage 3 = 64/81 × 8/9 = 512/729.
3. Stage 4 = 512/729 × 8/9 = 4096/6561. Stage 5 = 4096/6561 × 8/9 = 32768/59049.
4. Explicit formula: red area at stage n = (8/9)ⁿ. Recursive: (area at stage n) = 8/9 × (area at stage n − 1), for n ≥ 1, with stage 0 = 1.
5. Since 0 < 8/9 < 1, repeatedly multiplying by 8/9 makes the area smaller and smaller, getting as close to 0 as we like, but it never actually becomes 0 or negative.

AnswerStages 1, 2, 3: 8/9, 64/81, 512/729 sq units. Stages 4, 5: 4096/6561, 32768/59049 sq units. Explicit: (8/9)ⁿ. Recursive: a_n = (8/9)·a_n−1 (n ≥ 1), a₀ = 1. As n increases, the red area approaches 0.

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1. Let the first term be a and the common difference be d. The 11th term is a + 10d = 38, and the 16th term is a + 15d = 73.
2. Subtract the first equation from the second: 5d = 35, so d = 7.
3. Put d = 7 back into a + 10d = 38: a = 38 − 70 = −32.
4. The 31st term is a + 30d = −32 + 30 × 7 = −32 + 210 = 178.

AnswerThe 31st term is 178.

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1. Let the first term be a and the common difference be d. The third term gives a + 2d = 16.
2. The 7th term exceeds the 5th term by 12, so (a + 6d) − (a + 4d) = 12, which simplifies to 2d = 12, so d = 6.
3. Substitute d = 6 into a + 2d = 16: a + 12 = 16, so a = 4.

AnswerThe AP is 4, 10, 16, 22, … (first term a = 4, common difference d = 6).

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1. The smallest three-digit multiple of 7 is 105 (since 7 × 15 = 105), and the largest is 994 (since 7 × 142 = 994).
2. These multiples of 7 form an AP with first term a = 105, common difference d = 7, and last term l = 994.
3. The number of terms is one more than the number of steps: n = (l − a)/d + 1 = (994 − 105)/7 + 1 = 889/7 + 1 = 127 + 1 = 128.

AnswerThere are 128 three-digit numbers divisible by 7.

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1. The smallest multiple of 4 greater than 10 is 12, and the largest multiple of 4 less than 250 is 248.
2. These multiples form an AP with first term a = 12, common difference d = 4, last term l = 248.
3. The number of terms is n = (l − a)/d + 1 = (248 − 12)/4 + 1 = 236/4 + 1 = 59 + 1 = 60.

AnswerThere are 60 multiples of 4 between 10 and 250.

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Open NCERT p. 195Checked by computerAnswers can differ: one example

1. Let the first term be a and the common ratio be r. In a GP, a ≠ 0 and r ≠ 0. The terms are a, ar, ar², ar³, ar⁴, …
2. The fifth term is 4 times the third term: ar⁴ = 4 × ar². Divide both sides by ar² (it is not 0): r² = 4, so r = 2 or r = −2.
3. The sum of the first two terms is −4: a + ar = −4, that is, a(1 + r) = −4.
4. Take r = −2: a(1 − 2) = −4, so −a = −4 and a = 4. The GP is 4, −8, 16, −32, 64, …
5. Check: 4 + (−8) = −4 ✓. The third term is 16 and the fifth term is 64, and 64 = 4 × 16 ✓.
6. Taking r = 2 instead gives a(1 + 2) = −4, so a = −4/3. That gives a second GP, −4/3, −8/3, −16/3, −32/3, −64/3, …, which also meets both conditions.

AnswerOne such GP is 4, −8, 16, −32, 64, … (first term 4, common ratio −2). The question asks for a GP, and there are exactly two. The other is −4/3, −8/3, −16/3, … (first term −4/3, common ratio 2).

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1. A single number on its own is not a sum, so we look for runs of 2 or more consecutive natural numbers.
2. Suppose the run has k numbers and starts at a: a, a + 1, …, a + (k − 1). Its sum is k × a + (1 + 2 + ⋯ + (k − 1)) = k × a + k(k − 1)/2.
3. Set this equal to 100: k × a = 100 − k(k − 1)/2, so a = [100 − k(k − 1)/2] ÷ k. For a real run, a must be a natural number.
4. The run cannot be very long. Even starting at 1, k numbers add up to k(k + 1)/2, and that is already more than 100 when k = 14 (14 × 15 ÷ 2 = 105). So we only need to try k = 2 to 13.
5. k = 2: 99 ÷ 2 ✗. k = 3: 97 ÷ 3 ✗. k = 4: 94 ÷ 4 ✗. k = 5: 90 ÷ 5 = 18 ✓. k = 6: 85 ÷ 6 ✗. k = 7: 79 ÷ 7 ✗. k = 8: 72 ÷ 8 = 9 ✓. k = 9: 64 ÷ 9 ✗. k = 10: 55 ÷ 10 ✗. k = 11: 45 ÷ 11 ✗. k = 12: 34 ÷ 12 ✗. k = 13: 22 ÷ 13 ✗.
6. k = 5, a = 18: 18 + 19 + 20 + 21 + 22 = 100. k = 8, a = 9: 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100.

AnswerThere are exactly two ways: 100 = 18 + 19 + 20 + 21 + 22, and 100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16.

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1. The count doubles every hour starting from 30, so after n hours the count is 30 × 2ⁿ (n = 0 gives back the original 30).
2. End of 2nd hour: 30 × 2² = 30 × 4 = 120.
3. End of 4th hour: 30 × 2⁴ = 30 × 16 = 480.

AnswerEnd of 2nd hour: 120 bacteria. End of 4th hour: 480 bacteria. End of nth hour: 30 × 2ⁿ bacteria.

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1. Let the first term be a and common difference d. The 4th and 8th terms sum to (a+3d) + (a+7d) = 2a + 10d = 24, so a + 5d = 12.
2. The 6th and 10th terms sum to (a+5d) + (a+9d) = 2a + 14d = 44, so a + 7d = 22.
3. Subtracting the two equations: 2d = 10, so d = 5. Then a = 12 − 25 = −13.
4. The first three terms are a, a+d, a+2d = −13, −8, −3.

AnswerThe first three terms are −13, −8, −3.

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1. The sum of the first n natural numbers is n(n+1)/2. We need n(n+1)/2 > 1000, i.e. n(n+1) > 2000.
2. Check n = 44: 44 × 45 = 1980, and 1980/2 = 990, which is not greater than 1000.
3. Check n = 45: 45 × 46 = 2070, and 2070/2 = 1035, which is greater than 1000.

AnswerThe smallest such n is 45 (the sum reaches 1035, the first total past 1000).

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1. The first term is a = 2, and the common ratio is r = 8 ÷ 2 = 4 (check: 32 ÷ 8 = 4 too).
2. Explicit formula: nth term = a·rⁿ⁻¹ = 2 × 4ⁿ⁻¹. Recursive formula: nth term = 4 × (previous term), for n ≥ 2, with first term 2.
3. Set 2 × 4ⁿ⁻¹ = 131072, so 4ⁿ⁻¹ = 65536.
4. Since 4⁸ = 65536, we get n − 1 = 8, so n = 9.

Answer131072 is the 9th term. Explicit formula: 2 × 4ⁿ⁻¹. Recursive formula: term_n = 4 × term_n−1 (n ≥ 2), term₁ = 2.

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Open NCERT p. 195Checked by computerAnswers can differ: one example

1. Write the three terms as a/r, a, ar. Each is r times the one before it, and this form makes the product simple.
2. Product: (a/r) × a × (ar) = a³ = −1, so a = −1.
3. Sum: −1/r − 1 − r = 13/12. Add 1 to both sides: −1/r − r = 25/12, so 1/r + r = −25/12.
4. Multiply both sides by 12r: 12 + 12r² = −25r, that is, 12r² + 25r + 12 = 0.
5. Factorise: 12r² + 25r + 12 = (4r + 3)(3r + 4), since (4r + 3)(3r + 4) = 12r² + 16r + 9r + 12. So r = −3/4 or r = −4/3.
6. For r = −3/4: a/r = −1 ÷ (−3/4) = 4/3, a = −1, ar = (−1) × (−3/4) = 3/4. The terms are 4/3, −1, 3/4.
7. For r = −4/3: the terms are 3/4, −1, 4/3, the same three numbers in reverse order.
8. Check: 4/3 − 1 + 3/4 = 16/12 − 12/12 + 9/12 = 13/12 ✓, and (4/3) × (−1) × (3/4) = −1 ✓.

AnswerThe common ratio is −3/4 and the terms are 4/3, −1, 3/4. With the ratio −4/3 you get the same three terms in reverse order: 3/4, −1, 4/3. Both answers are correct.

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Open NCERT p. 195One way to think about it

1. Let the GP have first term a and common ratio r. In a GP, a ≠ 0 and r ≠ 0, so no term is 0. The nth term is a·rⁿ⁻¹.
2. So x = 4th term = a·r³, y = 10th term = a·r⁹, and z = 16th term = a·r¹⁵.
3. Divide each by the one before it: y ÷ x = a·r⁹ ÷ a·r³ = r⁶, and z ÷ y = a·r¹⁵ ÷ a·r⁹ = r⁶.
4. Both ratios are the same number, r⁶. So x, y, z have a common ratio, which means they are in GP.
5. This works because 4, 10, 16 are equally spaced: each jump is 6 places, which is 6 multiplications by r. The same fact written another way: y² = a²r¹⁸ = (a·r³)(a·r¹⁵) = xz.

In shorty ÷ x = z ÷ y = r⁶, so x, y, z are in GP with common ratio r⁶.

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1. Write the three terms as a/r, a, ar. Let S = 1/r + 1 + r. The sum condition gives a × S = 26.
2. The sum of the squares is a²/r² + a² + a²r² = a²(1/r² + 1 + r²) = 364.
3. Square S: (1/r + 1 + r)² = 1/r² + 1 + r² + 2(1/r + r + 1). So 1/r² + 1 + r² = S² − 2S.
4. So a²(S² − 2S) = 364, that is, (aS)² − 2a(aS) = 364. Put aS = 26: 676 − 52a = 364. So 52a = 312 and a = 6.
5. Then S = 26 ÷ 6 = 13/3, so 1/r + 1 + r = 13/3 and 1/r + r = 10/3. Multiply both sides by 3r: 3 + 3r² = 10r, that is, 3r² − 10r + 3 = 0.
6. Factorise: (3r − 1)(r − 3) = 0, because (3r − 1)(r − 3) = 3r² − 9r − r + 3. So r = 3 or r = 1/3.
7. For r = 3 the terms a/r, a, ar are 2, 6, 18. For r = 1/3 they are 18, 6, 2, the same numbers in reverse order.
8. Check: 2 + 6 + 18 = 26 ✓, and 2² + 6² + 18² = 4 + 36 + 324 = 364 ✓.

AnswerThe terms of the GP are 2, 6, 18 (common ratio 3). With common ratio 1/3 the same terms come in reverse order: 18, 6, 2.

Watch the lesson Common ratio, and why the nth term is ar^(n−1) · हिंदी में देखें

Open NCERT p. 195Checked by computer

1. P₁ = 1 and P₂ = 2 are given. For n > 2, add all the earlier terms and then add 1.
2. P₃ = 1 + 2 + 1 = 4. P₄ = 1 + 2 + 4 + 1 = 8. P₅ = 1 + 2 + 4 + 8 + 1 = 16. P₆ = 1 + 2 + 4 + 8 + 16 + 1 = 32. P₇ = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64. P₈ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128.
3. A simpler rule: for n ≥ 4, the given rule for P_n−1 says P_n−1 = P₁ + ⋯ + P_n−2 + 1. So P_n = (P₁ + ⋯ + P_n−2 + 1) + P_n−1 = P_n−1 + P_n−1 = 2P_n−1.
4. It also holds for n = 2 and n = 3: P₂ = 2 = 2 × P₁ and P₃ = 4 = 2 × P₂. So P_n = 2P_n−1 for every n ≥ 2, with P₁ = 1.
5. Each term is double the one before, so this is a GP with first term 1 and common ratio 2. Explicit formula: P_n = 1 × 2ⁿ⁻¹ = 2ⁿ⁻¹.

AnswerP₁ to P₈: 1, 2, 4, 8, 16, 32, 64, 128. Simpler recursive formula: P_n = 2P_n−1 for n ≥ 2, with P₁ = 1. Explicit formula: P_n = 2ⁿ⁻¹.

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Open NCERT p. 195Checked by computer

1. W₁ = 1 and W₂ = 2 are given. For n > 2, add the terms from W₁ up to W_n−2 (stopping two places back), then add 2.
2. W₃ = W₁ + 2 = 1 + 2 = 3. W₄ = 1 + 2 + 2 = 5. W₅ = 1 + 2 + 3 + 2 = 8. W₆ = 1 + 2 + 3 + 5 + 2 = 13. W₇ = 1 + 2 + 3 + 5 + 8 + 2 = 21. W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34.
3. Why each term is the sum of the two before it: for n ≥ 4, W_n = (W₁ + ⋯ + W_n−3) + W_n−2 + 2, and the rule for W_n−1 says W_n−1 = (W₁ + ⋯ + W_n−3) + 2.
4. Subtract: W_n − W_n−1 = W_n−2, so W_n = W_n−1 + W_n−2. It also holds for n = 3: W₃ = 3 = W₂ + W₁.
5. So W starts 1, 2 and each term from the third on is the sum of the two before it. That is exactly the Virahānka–Fibonacci sequence defined in this chapter: V₁ = 1, V₂ = 2, V_n = V_n−1 + V_n−2 for n ≥ 3.

AnswerW1 to W8 are 1, 2, 3, 5, 8, 13, 21, 34. This is the Virahānka–Fibonacci sequence from this chapter (V1 = 1, V2 = 2, and each term from the third on is the sum of the two before it).

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