Sequences and Series · Lesson 4 of 4
Chapter Summary and Practice
“Connect the chapter’s rules, representations and models, then use them in mixed reasoning.”
• Connect explicit and recursive sequence rules to functions and finite series. • Choose correctly between a term, a partial sum, a geometric mean and a product. • Explain the derivations and restrictions of the chapter’s main formulas. • Use geometric structure to connect selected terms, blocks, products and reciprocal sums. • Distinguish arithmetic and geometric change in mixed practical problems. • Check candidate solutions, sign branches and indexing through representative chapter problems.
One chapter, several ways to describe change
The chapter began with ordered observations and developed rules for working with them. A sequence tells us what occupies each position. A series tells us which terms are being added. A geometric progression is a special sequence whose repeated multiplication makes distant terms and finite sums predictable. This lesson connects those ideas instead of treating the formulas as separate facts.
Think first about the question being asked. “What is the tenth value?” concerns one term. “How much altogether through ten positions?” concerns a sum. “Which position has this value?” asks for an index. “What can be inserted to balance the ratios?” asks for geometric means. Each question can use the same sequence while requiring a different calculation.
| Chapter topic | Central meaning | Method or connection |
|---|---|---|
| Sequence and position | Each natural-number input selects one term. | aₙ or a(n); finite domains stop at a stated position. |
| Explicit and recursive rules | A direct formula uses the index; a recurrence uses earlier terms. | Substitute an index, or begin with enough initial values. |
| Series and partial sum | Addition signs form a series; Sₙ is its finite total. | Read sigma limits carefully and include the correct positions. |
| Geometric progression | Every nonzero term is r times the preceding term. | Check consecutive ratios; aₙ = arⁿ⁻¹. |
| Finite geometric sum | Multiplication shifts the sum and subtraction cancels interior terms. | Use the fraction formula for r ≠ 1, and na for r = 1. |
| Means and insertions | Equal multiplicative steps connect positive endpoints. | G = √ab; n insertions create n + 1 gaps. |
| Arithmetic and geometric means | A sum and a product encode a pair of positive numbers. | A ≥ G, equality exactly for equal numbers; recover A ± √(A² − G²). |
| Selected terms and blocks | Equal gaps or equal block lengths give the same multiplicative shift. | Use powers rᵈ or rⁿ; check nonzero sums before division. |
| Products and reciprocals | Corresponding terms combine or reverse ratio factors. | Ratios rR and 1/r; paired endpoints give P² = (ab)ⁿ. |
| Algebraic proofs | Factorisation or zero squares reveal hidden equal ratios. | Justify cancellations and nonzero conditions. |
| Related series and models | A non-geometric list can contain a geometric expression. | Rewrite repeated digits; distinguish additions from percentage updates. |
Reconstruct the formulas from their meanings
A formula is easier to remember when its structure answers a geometric or counting question. The exponent in a term counts the number of ratio steps. The exponent in a sum comes from the additional term in the shifted list. An inserted mean occupies an intermediate position, so the number of gaps differs from the number of inserted entries.
For a G.P. with first term a and ratio r, reaching position n takes n − 1 multiplication steps. For its first-n-term sum, multiplying by r creates the extra final term arⁿ. Subtracting the two sum expressions leaves a − arⁿ; that is why the sum formula contains exponent n rather than n − 1.
For positive endpoints u and v, one geometric mean is the positive value √uv. If n means are inserted, there are n + 2 entries altogether and n + 1 ratio steps. A problem asking for arbitrary real G.P. insertions may allow an additional negative-ratio branch; a problem asking for positive means does not.
The arithmetic mean A = (u + v)/2 gives the sum, while G = √uv gives the product. The nonnegative square (√u − √v)² establishes A ≥ G. Conversely, u and v can be reconstructed from the centre A and the displacement √(A² − G²). These ideas link averaging to quadratic equations.
Choose a representation that fits the information
The same G.P. can be written with the first term, a middle term or its endpoints as the starting information. Choose the representation that simplifies the given conditions, and then explain what your symbols mean. A convenient choice saves algebra without removing the need for reasoning.
| Question or condition | Useful starting point | Essential check |
|---|---|---|
| A specified position | aₙ = arⁿ⁻¹ | Count from position 1. |
| A specified value | Set arⁿ⁻¹ equal to that value. | The resulting n must be a positive integer. |
| Two distant terms | Divide the later-term equation by the earlier. | An even exponent can permit both signs of r. |
| A finite total | Write Sₙ, with a separate r = 1 case. | Distinguish n included terms from the nth term. |
| Three terms and their product | m/r, m, mr | m names the middle term. |
| Repeated digits or decimals | Write each entry using powers of 10. | The original list need not itself be geometric. |
| Percentage change over time | Initial value × update factorᵗ | Fix time zero and count completed updates. |
| Equal principal instalments | List starting unpaid balances and interest charges. | Specify when interest is charged relative to repayment. |
Problem
Given a₁ = 3 and aₙ = 2aₙ₋₁ for n ≥ 2, find a₆ and S₆, explaining why their formulas differ.
- 1.The recurrence doubles each preceding term, so the G.P. has a = 3 and r = 2.
- 2.The direct rule is aₙ = 3 × 2ⁿ⁻¹. Therefore a₆ = 3 × 2⁵ = 96.
- 3.The finite series is 3 + 6 + 12 + 24 + 48 + 96.
- 4.Use S₆ = 3(2⁶ − 1)/(2 − 1) = 3 × 63 = 189.
- 5.The term calculation uses five multiplication gaps; the sum calculation contains 2⁶ because multiplying the sum by 2 creates a seventh-position term.
- 6.Both numbers describe the same sequence, but 96 is one value and 189 is a total.
Problem
The second term is 6 and the fourth is 24. Find every real G.P. fitting these facts and the first-four-term sum for each.
- 1.The equations are ar = 6 and ar³ = 24. Dividing gives r² = 4.
- 2.For r = 2, a = 3, so the first four terms are 3, 6, 12, 24 and S₄ = 45.
- 3.For r = −2, a = −3, so the first four terms are −3, 6, −12, 24.
- 4.Its sum is S₄ = −3[1 − (−2)⁴]/[1 − (−2)] = −3(−15)/3 = 15.
- 5.Both have the required second and fourth values. The sum distinguishes them even though the supplied even-gap term information does not.
Problem
Insert two positive means between 2 and 54, then compare the arithmetic and geometric means of the endpoints.
- 1.The complete list has four terms and three gaps, so 54 = 2r³.
- 2.Thus r³ = 27 and the positive ratio is r = 3. The list is 2, 6, 18, 54.
- 3.The endpoint arithmetic mean is A = (2 + 54)/2 = 28.
- 4.The endpoint geometric mean is G = √108 = 6√3, approximately 10.392.
- 5.Therefore A > G because the endpoints differ.
- 6.The two inserted means 6 and 18 are not each equal to the single geometric mean of the endpoints; they solve a different gap-count problem.
Problem
Find the sum of the first three entries 8, 88, 888, … using the transformed-series method.
- 1.The kth entry is (8/9)(10ᵏ − 1), so the three-term total is (8/9)[10 + 100 + 1000 − 3].
- 2.The bracket is 1110 − 3 = 1107.
- 3.Thus the total is (8/9) × 1107 = 8 × 123 = 984.
- 4.Checking directly gives 8 + 88 + 888 = 984. The transformation works because a geometric sum is hidden inside each repeated-digit expression.
Problem
Three nonzero G.P. terms sum to 13 and have product 27. Find the possible lists.
- 1.Represent the terms as m/r, m, mr. Their product is m³ = 27, so m = 3.
- 2.The sum equation is 3/r + 3 + 3r = 13.
- 3.Multiply by r: 3r² − 10r + 3 = 0.
- 4.Factor (3r − 1)(r − 3) = 0, so r = 1/3 or r = 3.
- 5.The lists are 9, 3, 1 and 1, 3, 9. Both sum to 13 and have product 27.
- 6.Reversing a G.P. replaces its ratio by its reciprocal, which explains the two ordered solutions.
Problem
A model starts with 50 units at time zero and multiplies the amount by 1.2 each hour. Find the amount after two hours and the sum of the observations at times 0, 1 and 2.
- 1.At time 2 the value is 50(1.2)² = 72 units.
- 2.The three observations are 50, 60 and 72, so their numerical sum is 182 units.
- 3.The same total follows from a three-term G.P. sum: 50[(1.2)³ − 1]/(1.2 − 1) = 182.
- 4.The sum of observations is a separate quantity from the current amount. If the observations describe the same stock at different times, adding them is not a claim that 182 units physically exist at time 2.
- 5.Always connect a mathematical sum to what the question actually intends to measure.
Connect the deeper relationships
Several chapter proofs depend on the same idea: moving a fixed number of positions multiplies a term by a fixed power of the ratio. This makes equally spaced terms, equal-length blocks and reversed endpoint pairs predictable. Product relationships and sums of reciprocals are consequences of that structure.
For equally spaced selected positions p, p + d and p + 2d, the selected ratio is rᵈ and the middle value squared equals the outer product. For two consecutive n-term blocks, the later sum is rⁿ times the earlier sum. For first and last values a and b, each oppositely placed pair has product ab, so the squared total product is (ab)ⁿ. These are three views of the same exponent-counting structure.
A proof based on sums of squares uses a different but equally reusable principle. Real squares are nonnegative; if their sum is at most zero, every one must vanish. The resulting equalities may force b = ar, c = ar² and d = ar³. Nonzero checks then establish the ratio-based definition. The proof should show these steps rather than simply announcing that the list is geometric.
Errors that a final check should catch
Many mistakes come from choosing a plausible formula before interpreting the question. A good check uses the original wording or condition, not just the transformed equation. Substitute the answer back, examine its sign and test a small position where the intended value is already known.
No. A definite recursive or verbal rule can also select its terms. A few displayed values alone do not determine a unique continuation.
Do not apply a G.P. formula merely because a problem contains a sequence. Check the ratio rule or rewrite the expression to expose a genuine geometric component. Also, finite partial sums do not establish that an infinite series has a finite total.
Chapter check
Work across the whole chapter: interpret the rule, select the representation and use restrictions to decide what counts as a valid answer. The questions below combine recall with reasoning rather than repeating one calculation pattern.
Quiz
For a₁ = 2 and aₙ = aₙ₋₁ + 3, which list gives the first four terms?
What is Σ from k = 1 to 3 of 2ᵏ?
Which list is a G.P. under the nonzero-term definition?
What is S₄ for a = 2 and r = 3?
A second term and a fourth term give r² = 9. Which conclusion is justified without more information?
Three positive means are inserted between endpoints. How many terms are in the full list?
Positive numbers have A.M. 5 and G.M. 4. What are the numbers?
A value starts at 100 at time zero and grows by 10% per period. Which gives its value after two periods?
For three equally spaced G.P. terms x, y, z, which identity is guaranteed?
The sum of the first n terms is S and the sum of the next n is T. Which holds even if S = 0?
Practice Problems
- For aₙ = (−1)ⁿ⁻¹(n + 1), list five terms, write their series and evaluate the sum. Explain why the list is not a G.P.
- Given a₁ = 2, a₂ = 3 and aₙ = aₙ₋₁ + aₙ₋₂ for n > 2, find six terms and explain why two initial values are necessary.
- Find the eighth term and first-eight-term sum of 3, 6, 12, … .
- Find all real G.P.s with third term 20 and fifth term 80; explain the ratio ambiguity.
- A G.P. has first term 5, ratio 2 and finite sum 315. Find its length and last term, then check both.
- Insert three positive means between 2 and 162, and investigate whether a negative-ratio real insertion also fits.
- Positive numbers have A.M. 10 and G.M. 6. Find them and construct their monic quadratic.
- Prove the endpoint-product identity for a general n-term G.P. without assuming all terms are positive.
- Two G.P.s have first terms 2 and 3 and ratios 4 and 1/2. Find the sum of their first five corresponding products.
- Derive the total of 0.3 + 0.33 + 0.333 + … through n entries.
- A machine worth ₹8000 loses 25% of its current value annually. Find its value after three years and compare it with fixed annual subtraction of ₹2000.
- Under starting-balance interest, a ₹3000 debt is repaid in five equal annual principal instalments with 10% interest. Find its total interest and identify the progression of interest charges.
- For four consecutive G.P. terms, prove (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)² by factoring a common ratio.
- A function on positive integers has f(x + y) = f(x)f(y) and f(1) = 2. Find the first five terms and their total.
- Explain the difference between summing stage counts in a branching process and summing repeated observations of the same changing stock.
Key Takeaways
• A sequence is a function of position; a series indicates addition of selected terms. • Use a direct index rule or a recurrence with enough initial values to generate terms. • A nonzero constant ratio creates a G.P.; count n − 1 gaps to reach its nth term. • Derive finite sums by shifting and subtracting, and treat r = 1 separately. • Positive geometric means balance equal ratios, while A.M. and G.M. encode a sum and a product. • Equally spaced terms, shifted blocks, corresponding products and reversed pairs all preserve predictable geometric structure. • Hidden geometric sums can solve repeated-digit and repeated-decimal totals even when the original list is not a G.P. • Practical models require clear time origins, update rules and payment timing. • A correct answer includes justified restrictions, all admissible sign branches and a check against the original conditions.
Previous · Lesson 3
Mixed Applications
Next
End of chapter