Geometric and Mixed Sequence Problems

Geometric and mixed sequence problems combine the formulas for geometric sequences with those for arithmetic sequences, and they are solved by translating each condition into an equation for a₁, r and d. In brief: if three numbers are in geometric progression with the product 64 and the sum 21, the middle term is 4 (b³ = 64), and the others are 4/r and 4r with 4/r + 4r = 17, so r = 4 or 1/4: the numbers are 1, 4, 16.

Typical problems ask for terms or ratios from given data, for sums with a condition, for sequences that are arithmetic in one respect and geometric in another and for applications such as loans, populations and repeated percentages. This guide gives a routine, worked solutions for a wide range of problem types and a table that links the problem type with the method.

How do you approach these problems?

1. Decide for each given sequence whether it is arithmetic (constant difference) or geometric (constant ratio).
2. Introduce unknowns, usually a₁ and r (or d).
3. Write one equation for each condition, using a_n = a₁rⁿ⁻¹, S_n = a₁(rⁿ − 1)/(r − 1) and the rule b² = ac for geometric neighbours.
4. Divide equations to eliminate a₁ when possible. Solve for r first.
5. Check all solutions against the conditions. Negative r and several roots are common.

Geometric sequence problems

Problem 1: from two terms

a₃ = 18 and a₆ = 486. Divide: r³ = 27, so r = 3. Then a₁ = 18/9 = 2, and the sequence is 2, 6, 18, 54, 162, 486, …

Problem 2: sums of terms

a₁ + a₂ = 12 and a₃ + a₄ = 108. The second sum equals r² times the first, so r² = 9 and r = 3 or r = −3. For r = 3: a₁(1 + 3) = 12, so a₁ = 3: the sequence 3, 9, 27, 81. For r = −3: a₁(1 − 3) = 12, so a₁ = −6: the sequence −6, 18, −54, 162. Check the second: 162 − 54 = 108 ✓.

Problem 3: the first term and the ratio from the sum

The first three terms add up to 52 and the second term is 12. With the ratio r: 12/r + 12 + 12r = 52, so 12r² − 40r + 12 = 0, that is 3r² − 10r + 3 = 0, giving r = 3 or r = 1/3. For r = 3: 4, 12, 36. For r = 1/3: 36, 12, 4. The quadratic is solved with the quadratic formula: D = 100 − 36 = 64, r = (10 ± 8)/6.

Problem 4: the number of terms

1 + 3 + 9 + … + 3ⁿ⁻¹ = 3280. S_n = (3ⁿ − 1)/2 = 3280, so 3ⁿ = 6561 = 3⁸ and n = 8.

Problem 5: the product of terms

The product of the first n terms of a geometric sequence with a₁ = 1 and r = 2 is 2^{0+1+…+(n−1)} = 2^n(n−1)/2. For n = 5: 2¹⁰ = 1,024. Check: 1 · 2 · 4 · 8 · 16 = 1,024 ✓.

Mixed arithmetic and geometric problems

Problem 6: three numbers

Three positive numbers form an arithmetic sequence with the sum 15. If 1, 4 and 19 are added to them respectively, the results form a geometric sequence. Let the numbers be 5 − d, 5, 5 + d. After the additions: 6 − d, 9 and 24 + d. For a geometric sequence, 9² = (6 − d)(24 + d), so 81 = 144 − 18d − d², which gives d² + 18d − 63 = 0, so (d + 21)(d − 3) = 0 and d = 3 or d = −21. With d = 3: the numbers are 2, 5, 8 and the new ones 3, 9, 27 (ratio 3) ✓. With d = −21 the numbers would be 26, 5, −16, and −16 is not positive. So the numbers are 2, 5, 8.

Problem 7: a geometric sequence that becomes arithmetic

Three numbers form a geometric sequence with the sum 14. If the middle number is increased by 1, the three numbers form an arithmetic sequence. Find them. Write the numbers as a, ar, ar². The sum gives a(1 + r + r²) = 14. The arithmetic condition is 2(ar + 1) = a + ar², so a(r − 1)² = 2. Divide the first equation by the second: (1 + r + r²)/(r − 1)² = 7, so 1 + r + r² = 7r² − 14r + 7, which gives 6r² − 15r + 6 = 0, or 2r² − 5r + 2 = 0. Then r = 2 or r = 1/2. For r = 2: a = 14/7 = 2 and the numbers are 2, 4, 8 (after the change: 2, 5, 8, with difference 3). For r = 1/2: a = 8 and the numbers are 8, 4, 2 (after the change: 8, 5, 2).

Problem 8: exponent and index

The exponents of a geometric sequence a_n = 2³ⁿ⁻¹, namely 2, 5, 8, … form an arithmetic sequence with d = 3. The logarithm of a geometric sequence is an arithmetic sequence: log₂ a_n = 3n − 1. This links the two types.

Applied problems

Problem 9: savings with interest

Every year, 1,200 dollars is deposited at the end of the year into an account that pays 4% yearly. After 10 deposits, the first has earned interest for 9 years. The total is 1200(1.04⁹ + … + 1.04 + 1) = 1200 · (1.04¹⁰ − 1)/0.04 = 1200 · 12.006 = 14,407.3 dollars.

Problem 10: depreciation

A machine worth 50,000 dollars loses 15% of its value each year. After n years: 50000 · 0.85ⁿ. When does it fall below 20,000? 0.85ⁿ < 0.4, so n > ln 0.4/ln 0.85 = (−0.9163)/(−0.1625) = 5.64. So after 6 years.

Problem 11: a drug

A patient takes 100 mg every day. Each day 60% of the drug in the body is eliminated before the next dose, so 40% remains. Just after the nth dose, the amount is 100(1 + 0.4 + 0.4² + … + 0.4ⁿ⁻¹) = 100(1 − 0.4ⁿ)/0.6. After 5 doses: 100 · (1 − 0.01024)/0.6 = 164.96 mg. The long-term level is 100/0.6 = 166.7 mg.

Problem 12: a square spiral

A square has a side of 8. Joining the midpoints of its sides forms a smaller square, and so on. The side lengths: 8, 4√2, 4, 2√2, … with r = √2/2, and the areas are 64, 32, 16, … (r = 1/2). The area of the 6th square is 64 · (1/2)⁵ = 2.

Choosing the method

Where do such problems come from?

Mixed problems about numbers that are in arithmetic and geometric progression at the same time appear in medieval Arabic and Indian texts and in the works of Fibonacci. The relation between arithmetic and geometric sequences through logarithms was the key idea behind Napier’s invention in 1614: to multiply numbers (geometric progression), one adds their logarithms (arithmetic progression).

Where are they used today?

Financial planning (annuities, loans, depreciation), pharmacology (repeated doses) and population models rely on geometric sequences, with sums to find totals. The relation between the two types underlies the use of logarithmic scales for data growing by constant factors. Computing the number of periods to reach a goal needs logarithms.

Which mistakes are common?

Losing the negative ratio

From r² = 9 both 3 and −3 are possible, unless the context excludes one.

Mixing up the roles

Arithmetic: 2b = a + c. Geometric: b² = ac. Do not use one for the other.

Using the exponent n instead of n − 1

a_n = a₁rⁿ⁻¹. Check with the first term.

Dividing by an expression that can be zero

When dividing equations to eliminate a₁, make sure that a₁ ≠ 0 and the divisor is not zero.