The third term of a G.P. be one-fourth of fifth term and their sum is $1\dfrac{1}{4}$. If each term of this G.P. is positive, find its 10^th term.

Let first term of G.P. be a and common ratio be r.

Given,

The third term of a G.P. be one-fourth of fifth term.

∴ a₃ = $\dfrac{1}{4}a_5$

⇒ ar² = $\dfrac{ar^4}{4}$

⇒ 4ar² = ar⁴

⇒ $\dfrac{\text{ar}^4}{\text{ar}^2}$ = 4

⇒ r² = 4

⇒ r = $\sqrt{4} = \pm 2$.

Since, all terms of G.P. are positive, then r = 2.

Given,

Sum of third and fifth term = $1\dfrac{1}{4} = \dfrac{5}{4}$.

⇒ a₃ + a₅ = $\dfrac{5}{4}$

⇒ ar² + ar⁴ = $\dfrac{5}{4}$

⇒ 4(ar² + ar⁴) = 5

⇒ 4a(2² + 2⁴) = 5

⇒ 4a(4 + 16) = 5

⇒ 4a × 20 = 5

⇒ a = $\dfrac{5}{20 \times 4} = \dfrac{5}{80} = \dfrac{1}{16}$.

⇒ a₁₀ = ar⁹ = $\dfrac{1}{16} \times 2^9 = \dfrac{1}{16} \times 512$ = 32.