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(A) Let $G_{1}, G_{2}, G_{3}$ be three numbers between $1$ and $256$ such that $1, G_{1}, G_{2}, G_{3}, 256$ is a $G.P.$Here,the first term $a = 1$ an

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(A) Let $G_{1}, G_{2}, G_{3}$ be three numbers between $1$ and $256$ such that $1, G_{1}, G_{2}, G_{3}, 256$ is a $G.P.$
Here,the first term $a = 1$ and the fifth term $a_{5} = ar^{4} = 256$.
Substituting $a = 1$,we get $r^{4} = 256$,which implies $r = \pm 4$.
Case $1$: If $r = 4$,the numbers are $G_{1} = 1 \times 4 = 4$,$G_{2} = 1 \times 4^{2} = 16$,and $G_{3} = 1 \times 4^{3} = 64$.
Case $2$: If $r = -4$,the numbers are $G_{1} = 1 \times (-4) = -4$,$G_{2} = 1 \times (-4)^{2} = 16$,and $G_{3} = 1 \times (-4)^{3} = -64$.
Thus,the three numbers can be $4, 16, 64$ or $-4, 16, -64$.

Insert three numbers between $1$ and $256$ so that the resulting sequence is a $G.P.$

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