5B - Geometric sequences

5B - Geometric sequences

Repeated multiplication produces a different kind of pattern from repeated addition. You will find geometric terms, recover possible ratios from term conditions, and use logarithms to locate the first term beyond a growth or decay threshold.

Common ratio and nth term

The sequence 8,4,2,1,…8,4,2,1,\ldots keeps halving. Will any term actually equal zero?

Every term is half the previous term. After three multiplications the value is 8(12)3=18(\tfrac12)^3=1; at position nn, there have been n−1n-1 multiplications. A finite number of halvings leaves a positive value, however small.

Geometric sequence

A geometric sequence is formed by multiplying each term by the same common ratio, rr, to get the next term. Its terms begin a,ar,ar2,…a,ar,ar^2,\ldots.

When a term is nonzero, divide the next term by it to find rr. For example, 3,6,12,24,…3,6,12,24,\ldots has r=2r=2. The arithmetic sequence 3,6,9,12,…3,6,9,12,\ldots has no constant ratio.

With a nonzero first term, a negative ratio changes the sign at every multiplication. Its magnitude determines whether the sizes grow, shrink or stay the same.

[DIAGRAM: asset_name: Positive and negative halving ratios; asset_slug: signed-geometric-terms; caption: A negative halving ratio alternates signs while the term magnitudes shrink; description: Two discrete term plots compare first term 8 with ratios one half and minus one half. The positive sequence is 8,4,2,1,one half,one quarter; the negative-ratio sequence is 8,minus4,2,minus1,one half,minus one quarter. Points are shown at integer positions only.]
A negative halving ratio alternates signs while the term magnitudes shrink

Geometric nth term

For a nonzero common ratio and a positive integer nn,

un=arn−1u_n=ar^{n-1}

If r=0r=0, write the first term as u1=au_1=a and all later terms as 00, without evaluating 000^0. If r=1r=1, every term is aa.

Not on the formulae sheet: learn it.

The ratio belongs inside brackets when it is negative.

Worked example 1: repeated halving

Question. Find the sixth term of 8,4,2,1,…8,4,2,1,\ldots.

Plan. Use the first term 88 and five factors of 12\tfrac12.

r=48=12next divided by previousu6=8(12)5five multiplications=14evaluate exactly\begin{aligned} r&=\frac48=\frac12 &&\text{next divided by previous}\\ u_6&=8\left(\frac12\right)^5 &&\text{five multiplications}\\ &=\frac14 &&\text{evaluate exactly} \end{aligned}

Answer. 14\tfrac14.

Check. Continuing the listed terms gives u5=12u_5=\tfrac12, then u6=14u_6=\tfrac14.

Change only the sign of the ratio. The magnitudes are unchanged, but the sixth term now has an odd number of negative factors.

Worked example 2: a negative ratio

Question. Find the sixth term of 8,−4,2,−1,…8,-4,2,-1,\ldots.

Plan. Raise the entire ratio −12-\tfrac12 to the fifth power.

r=−48=−12retain the signu6=8(−12)5five signed factors=−14odd power is negative\begin{aligned} r&=\frac{-4}{8}=-\frac12 &&\text{retain the sign}\\ u_6&=8\left(-\frac12\right)^5 &&\text{five signed factors}\\ &=-\frac14 &&\text{odd power is negative} \end{aligned}

Answer. −14-\tfrac14.

Check. The fourth term is −1-1, the fifth is 12\tfrac12, and the sixth is −14-\tfrac14, as the plot shows.

For a nonzero first term, r=−1r=-1 alternates between aa and −a-a without changing size. If every term is zero, many ratios generate the sequence, so dividing terms cannot identify a unique ratio.

Finding geometric sequence parameters

Dividing two nonzero term equations removes the first term and leaves a power of the common ratio.

If up=arp−1u_p=ar^{p-1} and uq=arq−1u_q=ar^{q-1}, then uq/up=rq−pu_q/u_p=r^{q-p}, provided the denominator is nonzero. The exponent counts the gaps. An even power can leave two real sign possibilities; an odd power fixes the sign.

Worked example 3: two possible signs

Question. A geometric sequence has third term 1212 and fifth term 4848. Find all possible real values of rr and the corresponding first term.

Plan. Divide the fifth-term equation by the third-term equation, then test both square roots.

ar2=12,ar4=48translate the positionsr2=4812=4divide the equationsr=2orr=−2two real rootsa=12r2=3same r2 in either case\begin{aligned} ar^2&=12,\qquad ar^4=48 &&\text{translate the positions}\\ r^2&=\frac{48}{12}=4 &&\text{divide the equations}\\ r&=2\quad\text{or}\quad r=-2 &&\text{two real roots}\\ a&=\frac{12}{r^2}=3 &&\text{same }r^2\text{ in either case} \end{aligned}

Answer. r=2r=2 or −2-2, with a=3a=3 in both cases.

Check. Both 3,6,12,24,483,6,12,24,48 and 3,−6,12,−24,483,-6,12,-24,48 meet the two stated conditions. They differ at the even positions.

Keep the third term, but use a sixth-term condition. There are now three gaps, so an odd power determines one real ratio.

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