Chapter 2 · Sequences · Concept 2
2-10 Sum of Geometric Sequence (2)
Now the sums are what you are given, and the sequence is what you must find. It looks like it needs heavy algebra. It does not — it needs one observation about chunks of terms, and after that every question on the page is the same question.
Guiding question
If you know two sums, can you rebuild the sequence?
Learning outcome. Find the first term and common ratio of a geometric sequence from its sums.
A block of terms behaves like a single term
Take any run of terms and multiply the whole run by rk. You get the run that sits k places further along.
a₃ + a₄ + a₅ = r2(a₁ + a₂ + a₃)
a₄ + a₅ + a₆ = r3(a₁ + a₂ + a₃)
Why? Because ar² + ar³ + ar⁴ = r²(a + ar + ar²). Every term in the second block is r² times the matching term in the first, so factoring r² out is all that happens. The power is the shift — how many places along you moved.
And the version you will use most:
S₆ = S₃ + r3S₃
The first six terms are the first three, plus the next three — and the next three are r³ times the first three.
Light up the first block, then the second, and compare their totals. The second is never a new number to work out — it is the first one multiplied by a power of r.
terms 1 to 3
add to 35
1. The sum of terms 1 to 3 is 35, and the sum of terms 3 to 5 is 140. Find the first term and the common ratio.
- 1
Write the first fact plainly.
a + ar + ar2 = 35Three terms, so three pieces. Nothing clever yet.
The blocks do not overlap
terms 1–3 and terms 4–6
The second block is the first one shifted three places, so it is simply r³ × (first block). Divide one by the other and you have r³ straight away.
One block sits inside the other
terms 1–3 and terms 1–6
Split the big one first: S₆ = S₃ + r³S₃. Now the two facts are about the same thing and you can substitute.
Watch the overlap. “Terms 3 to 5” and “terms 1 to 3” share the 3rd term, but that does not matter — the shift is still 2, because term 3 is two places after term 1. Count the shift between the starting points, not the terms in common.
Work with a partner. For each one, first decide out loud whether the blocks overlap or sit apart. That decision picks the method, and then the algebra is short.
Find the first term and the common ratio.
Number 8 is worth a second look. Its quadratic turns out to be (r + 2)² = 0 — a perfect square, so the two roots are the same one. That is why it has a single answer even though the power was even.
For a geometric sequence, S₈ = 255, S₆ = 63 and a₈ − a₇ = 64. Find a₁₀.
How does changing r from a number bigger than 1 to a number between 0 and 1 change what a geometric sequence describes in the real world?
Think about what each one does over a long time.
Getting the shift wrong.
Terms 3–5 against terms 1–3 is a shift of 2, so the factor is r². Terms 4–6 is a shift of 3, so r³. Count from starting point to starting point.
Forgetting the second root again.
r² = 4 still means r = ±2, and each ratio gets its own first term. This lesson does not make that go away.
Discarding an untidy answer.
a = 35/3 and a = 78/7 look wrong and are not. Check them by adding the terms rather than by how they look.
Treating S₆ as if it excluded the first three terms.
S₆ means terms 1 to 6, so it CONTAINS S₃. Split it as S₃ + r³S₃ rather than comparing it directly.
Hunting for a when the question never asked.
The challenge needs only r and one term. Work out what the question actually wants before starting.
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