If we insert two numbers between sqrt{2} and | vedclass

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(B) Let the sequence be $\sqrt{2}, g_1, g_2, 4$. Here, the first term $a = \sqrt{2}$ and the fourth term $ar^3 = 4$. $r^3 = \frac{4}{\sqrt{2}} = \frac

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$2, \sqrt{8}$

Vedclass · Answer

(B) Let the sequence be $\sqrt{2}, g_1, g_2, 4$. Here, the first term $a = \sqrt{2}$ and the fourth term $ar^3 = 4$. $r^3 = \frac{4}{\sqrt{2}} = \frac{2^2}{2^{1/2}} = 2^{2 - 1/2} = 2^{3/2}$. Taking the cube root on both sides, $r = (2^{3/2})^{1/3} = 2^{1/2} = \sqrt{2}$. Thus, the inserted numbers are: $g_1 = a \cdot r = \sqrt{2} \cdot \sqrt{2} = 2$. $g_2 = a \cdot r^2 = \sqrt{2} \cdot (\sqrt{2})^2 = \sqrt{2} \cdot 2 = 2\sqrt{2} = \sqrt{8}$.

If we insert two numbers between $\sqrt{2}$ and $4$ so that the resulting sequence is in $G.P.$, then the inserted numbers in the order are

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