P2 4.4 - Geometric sequences and finite sums

Nth terms and common ratio

How many terms of $6+9+13.5+\cdots$ are needed before the total first exceeds $130$? The terms increase by multiplication, so adding a fixed difference will not answer this question. First we need a language for that multiplication.

In $5,15,45,135,\ldots$, each term is three times the previous term. Four terms require only three multiplications: this is the reason for the $n-1$ in the term formula.

Starting with first term $a$, the terms are $a,ar,ar^2,\ldots$. The exponent counts how many times you have multiplied after the first term.

Here $n$ is a positive integer. If $r=0$, write $u_1=a$ and all later terms as zero directly; there is no need to interpret $0^0$. A ratio cannot be recovered by dividing by a zero term.

Change only the sign of the common ratio. The magnitudes follow the same pattern, but the signs alternate.

A ratio compares in a particular direction.

Use the sign of the ratio as well as its magnitude.

Deriving the finite-sum formula

Multiplying a geometric sum by its ratio shifts every term one place along the pattern. Subtracting then cancels all the shared terms.

For the four-term sum $S_4=3+6+12+24$, multiply by $2$: $2S_4=6+12+24+48$. The middle three terms cancel when the first line is subtracted from the second, leaving $S_4=48-3=45$.

[DIAGRAM: asset_name: Cancelling a geometric sum; asset_slug: finite-gp-cancellation; caption: Multiplication by the ratio aligns all but the first and final terms.; description: Two aligned rows show S4 equal to 3 plus 6 plus 12 plus 24 and 2S4 equal to 6 plus 12 plus 24 plus 48. The matching 6, 12 and 24 terms are paired, leaving 2S4 minus S4 equal to 48 minus 3.]

The same cancellation proves the general result. Write exactly $n$ terms in each line:

Subtract the second line from the first:

The final step divides by $1-r$, so it requires $r\ne1$. You should be able to reproduce this proof, including the shifted line and cancellation.

For $r>1$, the second form avoids two negative quantities. For negative ratios, either form works if powers and brackets are handled carefully.

Change only the ratio from $2$ to $-2$. The sum formula still works, but the signs now matter.

The last term uses $r^{n-1}$, while the sum formula uses $r^n$: that extra power comes from the shifted row in the proof.

Finding the number of terms using logarithms

When the unknown is a term number, rearrange until a single power remains. Logarithms then bring its exponent down.

For our opening series, $a=6$ and $r=1.5$, so

This gives a compact equation for the whole total. Taking logs before isolating $1.5^n$ would leave a subtraction inside the logarithm; there is no log law that splits a difference.

Change the demand from an exact total to the first total exceeding a threshold. Now the logarithm gives a boundary, and $n$ must be an integer beyond it.

A question about the nth term instead begins with $u_n=ar^{n-1}$, not $S_n$. For a decreasing positive sequence, solving $ar^{n-1}<L$ may require division by $\log r<0$, which reverses the inequality. In either case, keep working precision and check neighbouring integer term numbers. A noninteger answer to an exact-total question may mean that no whole number of terms gives that total.

Quick Recap

• A common multiplier gives $u_n=ar^{n-1}$.
• Multiply the finite sum by $r$ and subtract to prove $S_n=a(1-r^n)/(1-r)$ for $r\ne1$.
• When $r=1$, use $S_n=na$.
• Isolate the power before taking logs, then interpret $n$ as an integer.

The opening total first exceeds $130$ after $7$ terms.