Check whether the sequence 5, 154, 4516, 13564, … is a geometric progression and find its nth term.
Answer: Every ratio is 34, so it is a GP with a = 5, r = 34. nth term = 5 × (34)n−1.
Step-by-step solution
Idea: Divide each term by the one before. If the answer is always the same, the sequence is a GP and tn = arn−1. Dividing by a fraction = multiplying by its reciprocal.
- t2t1 = 154 ÷ 5 = 154 × 15 = 34.½ mark
- t3t2 = 4516 ÷ 154 = 4516 × 415 = 34.½ mark
- t4t3 = 13564 ÷ 4516 = 13564 × 1645 = 34.½ mark
- The ratio is a constant 34, so the sequence is a GP with a = 5 and r = 34.½ mark
- tn = a × rn−1 = 5 × (34)n−1.1 mark
It is a GP with a = 5 and r = 3/4; nth term = 5 × (3/4)ⁿ⁻¹.
Check: n = 3: 5 × 916 = 4516 ✓; n = 4: 5 × 2764 = 13564 ✓.
Answer to write in the exam
t2t1 = 154 × 15 = 34
t3t2 = 4516 × 415 = 34
t4t3 = 13564 × 1645 = 34
Ratio is constant ∴ GP with a = 5, r = 34
tn = arn−1
∴ tn = 5 × (34)n−1
Common mistakes that cost marks
- Subtracting terms (checking for an AP) instead of dividing.
- Dividing fractions by multiplying straight across: 13564 ÷ 4516 written as 135 × 4564 × 16. Multiply by the reciprocal: 13564 × 1645 = 34.
- Writing the nth term as (5 × 34)n−1 = (154)n−1. Only r is raised to the power; 5 stays outside.
How this can come in the exam
MCQ (1 mark)
The common ratio of the GP 8, 6, 92, 278, … is
- 43
- 34
- −2
- 12
Show answer
(B) 34
68 = 34.
Short answer (2 marks)
Check whether 6, 125, 2425, … is a GP and write its nth term.
Show answer
125 ÷ 6 = 25, 2425 ÷ 125 = 25: a GP with r = 25 (1 mark). nth term = 6 × (25)n−1 (1 mark).Try one yourself
Check whether 9, 6, 4, 83, … is a GP and find its nth term.
Show answer
69 = 46 = 83 ÷ 4 = 23. GP with a = 9, r = 23; nth term = 9 × (23)n−1.
More questions like this
- Check whether the following sequences are geometric progressions and find their nth terms.
- Can you find a recursive rule for the formula tn = 3 × 10n−1 that generates the geometric progression 3, 30, 300, 3000, … ?
- Observe the Sierpiński triangle and try to answer the following questions
- The number of black triangles at the nth stage of the Sierpiński triangle is given by 3n. The number of black triangles increases very quickly as the stage numbers increase. Can you explain why?
- If the black region at Stage 0 is 1 square unit, then the black region at Stage 1 is 34 square units. This process is repeated on Stage 1 to arrive at Stage 2. Hence, the black region at Stage 2 will be 34 of the black region of Stage 1, which is equal to 34 × 34 = (34)2. Can you explain why the area of the black region at Stage n will be (34)n?
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