Predicting What Comes Next: Exploring Sequences and Progressions

Substitute n = 1, 2, 3, 4, 5 in each formula.

For 3n − 4: −1, 2, 5, 8, 11. For 2 − 5n: −3, −8, −13, −18, −23.

For n² − 2n + 3: 2, 3, 6, 11, 18.

t₁₀ = 5(10) − 3; t₁₅ = 5(15) − 3.

Solve 5n − 3 = 97: 5n = 100.

Solve 5n − 3 = 172: 5n = 175.

Solve 5n − 3 = 607 to obtain 5n = 610.

Start at −5 and add 3 repeatedly: −5, −2, 1, 4, 7.

The explicit form is −5 + 3(n − 1) = 3n − 8. For 52, solve 3n − 8 = 52.

Begin with 1, 2, 4. Each new term is the sum of the preceding three.

Then 1+2+4=7; 2+4+7=13; 4+7+13=24; 7+13+24=44; 13+24+44=81.

a = 3, d = 5, so tₙ = 3 + 5(n − 1) = 5n − 2.

Substitute n = 10 and 26.

The sequence starts at 21 with difference −3: tₙ = 24 − 3n.

For −81, n = (24 + 81)/3 = 35. For zero, n = 24/3 = 8.

The first term is 11 and the common difference is −3.

tₙ = 11 − 3(n − 1). A recursive definition must include the starting term.

a + 2d = 12 and a + 49d = 106. Subtract: 47d = 94, giving d = 2.

Then a = 8 and t₂₉ = 8 + 28(2).

The AP is 12, 15, …, 99, with difference 3.

n = (99 − 12)/3 + 1 = 30. Its sum is 30(12 + 99)/2.

The salary after k yearly increases is 500000 + 20000k.

Set this equal to 700000 to get k = 10. Starting salary is year 1, so ten increases lead to the salary of year 11.

The nth triangular number is 1 + 2 + … + n = n(n + 1)/2.

For n = 25, compute 25 × 26/2.

The eighth term is 192 and the common ratio is 2. Moving from term 8 to term 12 takes four multiplications by 2.

t₁₂ = t₈ × 2⁴ = 192 × 16. Equivalently the first term is 192/2⁷ = 3/2.

The sequence 5, 25, 125, … has first term 5 and ratio 5.

tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ.

Given t₁ = 2 and tₙ = 3tₙ₋₁ − 2, put uₙ = tₙ − 1. Then u₁ = 1 and uₙ = 3uₙ₋₁.

Thus tₙ = 3ⁿ⁻¹ + 1. For 730, solve 3ⁿ⁻¹ = 729 = 3⁶.

The sequence starts at 2 and multiplies by 3: tₙ = 2 × 3ⁿ⁻¹.

4374/2 = 2187 = 3⁷, so n − 1 = 7.

Recursive rule: t₁ = 2 and tₙ = 3tₙ₋₁ for n ≥ 2.

Each rebound reaches 60% of the preceding height. The fifth rebound is 80(0.6)⁵ = 6.2208 m.

To reach the sixth ground hit, the ball first falls 80 m and then traverses each of the first five rebound heights twice.

The first five heights sum to 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688. Total distance = 80 + 2(110.6688).

The printed sequence is 2, 2√2, 4, … . Its ratio is √2 and tₙ = 2(√2)ⁿ⁻¹ = 2⁽ⁿ⁺¹⁾/².

Since 128 = 2⁷, solve (n + 1)/2 = 7.

Stage 0 has one square. Each step replaces every retained square by eight smaller squares: counts at stages 0–3 are 1, 8, 64, 512.

Stages 4 and 5 contain 4096 and 32768 squares. Explicit: Nₙ = 8ⁿ. Recursive: N₀ = 1 and Nₙ = 8Nₙ₋₁.

Each step retains 8/9 of the previous area. A₀ = 1, Aₙ = (8/9)ⁿ, or recursively Aₙ = (8/9)Aₙ₋₁.

Areas at stages 1–5 are 8/9, 64/81, 512/729, 4096/6561, 32768/59049. These decrease towards zero.

a + 10d = 38 and a + 15d = 73. Subtraction gives 5d = 35, so d = 7.

a = 38 − 70 = −32. Then t₃₁ = −32 + 30(7).

t₇ − t₅ = 2d = 12, giving d = 6.

t₃ = a + 2d = 16 gives a = 4.

The first such multiple is 105 and the last is 994.

Count = (994 − 105)/7 + 1 = 128.

The multiples from 10 to 250 run from 12 to 248.

Count = (248 − 12)/4 + 1.

Let the first two terms be a and ar. Their sum is a(1 + r) = −4.

The fifth term equals four times the third: ar⁴ = 4ar². The nonzero-term interpretation gives r² = 4.

If r = 2, a = −4/3. If r = −2, a = 4. If zero ratios are allowed, r = 0 and a = −4 also satisfy the literal conditions, giving −4,0,0,… .

For k consecutive terms starting at positive integer a, 100 = k(2a + k − 1)/2. Hence a = (200/k − k + 1)/2.

Require a ≥ 1 and an integer. Since k(k + 1)/2 ≤ 100, only k ≤ 13 need checking; among divisors of 200 in that range, k = 1, 5 and 8 work.

Starting with 30, multiply by 2 for each elapsed hour.

After 2 hours: 30 × 2². After 4 hours: 30 × 2⁴.

t₄ + t₈ = 2a + 10d = 24. Also t₆ + t₁₀ = 2a + 14d = 44.

Subtract to get 4d = 20, so d = 5. Then 2a = 24 − 50 = −26.

Sₙ = n(n + 1)/2. S₄₄ = 990 and S₄₅ = 1035.

The sum increases with n, so 45 is the first index with sum greater than 1000.

For 2, 8, 32, …, tₙ = 2 × 4ⁿ⁻¹ = 2²ⁿ⁻¹.

131072 = 2¹⁷, so 2n − 1 = 17.

Explicit: tₙ = 2 × 4ⁿ⁻¹. Recursive: t₁ = 2, tₙ = 4tₙ₋₁ for n ≥ 2.

Write the terms as b/r, b, br. Their product is b³ = −1, giving b = −1.

Their sum is −1/r − 1 − r = 13/12. Thus 12r² + 25r + 12 = 0 = (3r + 4)(4r + 3).

The ratios are −4/3 and −3/4.

Let t₄ = x, t₁₀ = y and t₁₆ = z. Six steps multiply by r⁶, so y = xr⁶ and z = xr¹².

Then y² = x²r¹² = xz. This is also true in the zero-term degenerate case.

Let the terms be A, b, C, with AC = b² and A + b + C = 26.

A² + b² + C² = (26 − b)² − 2b² + b² = 676 − 52b = 364. Hence b = 6.

A + C = 20 and AC = 36. The roots of u² − 20u + 36 are 2 and 18.

Start with P₁ = 1 and P₂ = 2. Each new term is one more than the sum of all preceding terms.

The resulting eight terms are 1,2,4,8,16,32,64,128. Subtract two consecutive recurrence equations to get Pₙ = 2Pₙ₋₁.

Start W₁ = 1, W₂ = 2. For n ≥ 3, Wₙ is 2 plus the sum through Wₙ₋₂.

Compute 1,2,3,5,8,13,21,34. Subtract successive equations to get Wₙ − Wₙ₋₁ = Wₙ₋₂.

Naturals continue 7,8,9; odd numbers 13,15,17; triangular numbers 28,36,45; squares 49,64,81.

A finite sequence can describe the seven daily temperatures in a week. Its order matters.

Six and seven triangular rows contain 21 and 28 dots. The first n odd numbers total n², because each new odd-sized border turns an (n−1)-square into an n-square.

The sequence 1,4,7,10,13 continues 16,19,22,25. Its first ten partial sums are 1,5,12,22,35,51,70,92,117,145.

Triangular terms 5–8 are 15,21,28,36.

Examples of different sequence rules: 2n (even numbers), 2ⁿ (powers), n² (squares), (−1)ⁿ (alternating signs), and Fibonacci numbers with each new term the sum of two preceding terms.

The nth odd number is 2n − 1: terms 53,108,1170 are 105,215,2339.

For tₙ = 5n − 2, the first six terms are 3,8,13,18,23,28; terms 100 and 1000 are 498 and 4998.

308 occurs at n = (308+2)/5 = 62; 473 at n = (473+2)/5 = 95.

Indices are positive integers. Squares have rule n²; primes can be generated by testing divisibility, and the next primes after 29 are 31,37,41. A finite list alone does not determine a unique infinite continuation.

The first three terms are −4,−1,2. Terms 12,18,50 are 29,47,143.

For 332, solve 3n − 7 = 332 to get n = 113. For 557, n = 188.

From 1,5,13,29, apply tₙ = 2tₙ₋₁ + 3 to get 61,125,253.

Solving gives tₙ = 2ⁿ⁺¹ − 3. A term of 133 would require 2ⁿ⁺¹ = 136, which is not a power of two.

Each next term is the sum of its two predecessors.

After 21 and 34, obtain 55, then 89.

The counts begin 1,5,9,13. Each step adds four, giving tₙ = 4n − 3.

Stages 5,6,10,11,12,20 contain 17,21,37,41,45,77 squares.

An AP has a constant difference between consecutive terms. Naturals, odds, 1,4,7,… and −7,−3,1,… satisfy this. Triangular numbers, squares, reciprocals and primes do not have a constant difference.

For 2,5,8,…: tₙ = 3n − 1. For −5,−1,3,…: tₙ = 4n − 9.

For 1/2,5/2,9/2,…: tₙ = 2n − 3/2. For 1.5,3.5,5.5,…: tₙ = 2n − 0.5.

Each recursive rule gives its first term and adds its corresponding common difference (3,4,2,2).

Plotting each AP’s pairs (n,tₙ) gives points on a straight line; its slope is the common difference.

Two copies of a triangular dot array form n rows of n+1 dots: 2Sₙ = n(n+1).

For a general AP, pair first and last terms, second and next-to-last terms, and so on: Sₙ = n(a+l)/2 = n[2a+(n−1)d]/2.

S₂₀ = 210, S₅₀ = 1275, S₁₀₀₀ = 500500. Triangular numbers 10,17,80 are 55,153,3240.

The square counts start 3,6,12,24. Each new stage doubles: tₙ = 3 × 2ⁿ⁻¹.

Stages 5,6,10,11,12,20 contain 48,96,1536,3072,6144,1572864.

A fixed multiplier defines this GP; a fixed increment defines an AP.

For 2,10,50,250,… the ratio is 5, so tₙ = 2 × 5ⁿ⁻¹.

For 4,8/3,16/9,32/27,… the ratio is 2/3, so tₙ = 4(2/3)ⁿ⁻¹.

For 3,−3/2,3/4,−3/8,… the ratio is −1/2, so tₙ = 3(−1/2)ⁿ⁻¹.

For 3,30,300,… a recursive rule is t₁ = 3, tₙ = 10tₙ₋₁.

Starting at stage 0, the number of retained triangles is Nₙ = 3ⁿ, with N₀ = 1 and Nₙ = 3Nₙ₋₁.

Stages 0–3 have counts 1,3,9,27; stages 4 and 5 have 81 and 243 triangles.

Each step retains 3/4 of the area: Aₙ = (3/4)ⁿ, A₀ = 1.

Stages 1–5 have areas 3/4,9/16,27/64,81/256,243/1024. As stages increase, the area tends to zero.

Each rebound reaches 3/4 of the preceding height. Heights are 18,13.5,10.125,7.59375,5.6953125 feet for the first five rebounds.

The sixth height is 4.271484375 ft; the seventh is 3.20361328125 ft. Keep full precision until the end.