Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.
Answer: r2 = 4, so r = 2 or −2. r = 2: −43, −83, −163, …; r = −2: 4, −8, 16, −32, … Both GPs work.
Step-by-step solution
Idea: Write the terms as a, ar, ar2, …. The condition on the 5th and 3rd terms gives r; the sum of the first two terms then gives a.
- 5th term = 4 × 3rd term: ar4 = 4ar2. Divide by ar2 (not zero in a GP): r2 = 4 ⇒ r = 2 or r = −2.1½ marks
- Sum of first two terms: a + ar = a(1 + r) = −4.½ mark
- r = 2: 3a = −4 ⇒ a = −43. GP: −43, −83, −163, −323, …1 mark
- r = −2: −a = −4 ⇒ a = 4. GP: 4, −8, 16, −32, 64, …1 mark
Two GPs satisfy the conditions: −4/3, −8/3, −16/3, … (r = 2) and 4, −8, 16, −32, … (r = −2).
Check: r = −2: 4 + (−8) = −4 ✓, 5th term 64 = 4 × 16 ✓. r = 2: −4/3 − 8/3 = −4 ✓, 5th term −64/3 = 4 × (−16/3) ✓.
Answer to write in the exam
ar4 = 4ar2 ⇒ r2 = 4 ⇒ r = ±2
a + ar = a(1 + r) = −4
r = 2: 3a = −4 ⇒ a = −43 ∴ GP: −43, −83, −163, …
r = −2: −a = −4 ⇒ a = 4 ∴ GP: 4, −8, 16, −32, …
Common mistakes that cost marks
- Taking only r = 2. r2 = 4 has two solutions, 2 and −2, and both give a valid GP.
- Writing the 5th term as ar5 and the 3rd as ar3. The nth term is arn−1.
- Getting r = 4 from ‘4 times’. The 5th and 3rd terms are two steps apart, so the factor 4 is r2.
How this can come in the exam
MCQ (1 mark)
In a GP the 6th term is 9 times the 4th term. The common ratio can be
- 9 only
- 3 only
- 3 or −3
- 13
Show answer
(C) 3 or −3
ar5 = 9ar3 ⇒ r2 = 9 ⇒ r = ±3.
Try one yourself
Find a GP whose first two terms add up to 12 and whose 4th term is 9 times its 2nd term.
Show answer
r2 = 9 ⇒ r = ±3. r = 3: 4a = 12, a = 3: 3, 9, 27, …. r = −3: −2a = 12, a = −6: −6, 18, −54, …
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