The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a₅ = a r^5–1 = a r⁴ = p … (i)

a₈ = a r^8–1 = a r⁷ = q … (ii)

a₁₁ = a r^11–1 = a r¹⁰ = s … (iii)

Dividing equation (ii) by (i), we get

\( r^3=\dfrac{q}{p}\) ...............(iv)

Dividing equation (iii) by (ii), we get

\( r^3=\dfrac{s}{q}\) ...............(v)

Equating the values  of r³ obtained in (iv) and (v), we get

\( \dfrac{q}{p}=\dfrac{s}{q}\)

q² = ps

Hence proved