Geometric Progressions — MATHS STD 11 Science — Question

Let G₁, G₂, G₃, G₄, G₅ be 5 geometric means between $\text{a}=\frac{32}{9}$ and $\text{b}=\frac{81}{2}$

Then, $\frac{32}{9}$, G₁, G₂, G₃, G₄, G₅ $\frac{81}{2}$ is a G.P. with $\text{a}=\frac{32}{9}$, $\text{b}=\frac{81}{2}$

$\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}$

$=\Bigg(\frac{\frac{81}{2}}{\frac{32}{9}}\Bigg)^{\frac{1}{5+1}}=\Big(\frac{81}{2}\times\frac{9}{32}\Big)^{\frac{1}{6}}=\Big(\frac{{3}^6}{{2}^6}\Big)=\frac32$

$\therefore\text{G}_1=\text{ar}=\frac{32}{9}\times\frac32=\frac{16}{3}$

$\text{G}_2=\text{ar}^2=\frac{32}{9}\times\frac{9}{4}=8$

$\text{G}_3=\text{ar}^3=\frac{32}{9}\times\frac{27}{8}=12$

$\text{G}_4=\text{ar}^4=\frac{32}{9}\times\frac{{3}^4}{{2}^4}=2\times9=18$

$\text{G}_5=\text{ar}^5=\frac{32}{9}\times\frac{{5}^5}{{2}^5}=27$

Hence, $\frac{16}{3},8,12,18,27$ are 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}.$