Porisma

Worked solution

Recognise a common ratio

For each source-table list, state the first term aa and common ratio rr: (a) 5,25,125,6255,25,125,625; (b) 27,9,3,1,1/3,1/9,1/2727,9,3,1,1/3,1/9,1/27; (c) 2,−6,18,−54,1622,-6,18,-54,162. Then decide whether (d) 2,4,6,10,142,4,6,10,14 is a GP.

CJC · Sequences and Series · Unnumbered GP table and question, page 14

Source table: growing positive terms

Answer

a=5,r=5a=5,r=5.

Divide neighbouring terms

a=5,r=25/5=5a=5,\qquad r=25/5=5

The source identifies the list as a GP; these fill its two table columns.

How did this part go?

Source table: shrinking positive terms

Answer

a=27,r=1/3a=27,r=1/3.

Divide in the direction of the sequence

a=27,r=9/27=13a=27,\qquad r=9/27=\frac13

The next term is one third of the preceding term.

How did this part go?

Source table: alternating terms

Answer

a=2,r=−3a=2,r=-3.

Keep the sign of the ratio

a=2,r=(−6)/2=−3a=2,\qquad r=(-6)/2=-3

A negative common ratio is allowed for nonzero GP terms.

How did this part go?

The unnumbered question

Answer

No; the sequence is not a GP.

Compare two consecutive ratios

42=2,64=32≠2\frac42=2,\qquad\frac64=\frac32\ne2

The Teacher answer provides a counterexample to a common ratio.

How did this part go?

Solution notes

The three table rows and following question are one unnumbered activity. The part labels organise the source asks; they are not printed marks.