Geometric progressions · 3 marks

If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Answer: x = ar³, y = ar⁹, z = ar¹⁵, so yx = zy = r⁶. The ratio is the same, so x, y, z are in GP (equivalently y² = xz).

Step-by-step solution

Idea: Three numbers are in GP when the second divided by the first equals the third divided by the second. Write x, y, z using arⁿ⁻¹.

Let the GP have first term a and ratio r (both non-zero). Then x = ar³, y = ar⁹, z = ar¹⁵.1 mark
yx = ar⁹ar³ = r⁶ and zy = ar¹⁵ar⁹ = r⁶.1 mark
The ratios are equal (both r⁶), so x, y, z are in GP with common ratio r⁶. Hence proved.1 mark

Since y/x = z/y = r⁶, the numbers x, y, z are in GP (with common ratio r⁶).

Check: Example: GP 1, 2, 4, …: x = 2³ = 8, y = 2⁹ = 512, z = 2¹⁵ = 32768; 512/8 = 64 = 32768/512 ✓.

Answer to write in the exam

Let first term = a, common ratio = r

x = ar³, y = ar⁹, z = ar¹⁵

yx = ar⁹ar³ = r⁶

zy = ar¹⁵ar⁹ = r⁶

yx = zy

∴ x, y, z are in GP. Hence proved.

Common mistakes that cost marks

Writing the 4th term as ar⁴. The nth term is arⁿ⁻¹, so the 4th is ar³.
Showing only that y − x = z − y fails. That would test for an AP; a GP needs equal ratios.
Dividing exponents (r⁹r³ = r³). When dividing powers, subtract the exponents: r⁶.

How this can come in the exam

Assertion–Reason (1 mark)

Assertion (A): The 2nd, 5th and 8th terms of any GP are themselves in GP.
Reason (R): Terms of a GP whose positions are equally spaced have equal ratios.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Show answer

(A) Both A and R are true, and R is the correct explanation of A.
ar⁴ar = ar⁷ar⁴ = r³; R explains A.

Try one yourself

If the 3rd, 7th and 11th terms of a GP are p, q, s, show that q² = ps.