- A1 and 64
- B4 and 16
- C2 and 16
- D8 and 16.
Solution:
Let the two G.M.s between 1 and 64 be G1 and G2.
Thus, 1, G1, G2 and 64 are in G.P.
$64=1\times\text{r}^3$
$\Rightarrow\text{r}=\sqrt[3]{64}$
$\Rightarrow\text{r}=4$
$\Rightarrow\text{G}_1=\text{ar}=1\times4=4$
And, $\text{G}_2=\text{ar}^2=1\times4^2=16$
Thus, 4 and 16 are the required G.M.s.
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