Four geometric means are inserted between 1-8 and 128 - Find the third geometric mean-

The correct option is B 81/8× r^5 = 128 → r^5 = 128 × 8 = 1024 → r = 4. Thus, the series would be 1/8, 1/2, 2, 8, 32, 128. The third geometric mean would be 8. ...

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Four geometric means are inserted between $\frac{1}{8}$ and 128. Find the third geometric mean?

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Solution

The correct option is D 8
$\frac{1}{8} \times r^{5} = 128 \to r^{5} = 128 \times 8 = 1024 \to r = 4.$
Thus, the series would be $\frac{1}{8}, \frac{1}{2}, 2, 8, 32, 128$ .
The third geometric mean would be 8.

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Four geometric means are inserted between 18 and 128. Find the third geometric mean?

The correct option is D 818×r5=128→r5=128×8=1024→r=4. Thus, the series would be 18,12,2,8,32,128. The third geometric mean would be 8.

Four geometric means are inserted between $\frac{1}{8}$ and 128. Find the third geometric mean?

The correct option is D 8
$\frac{1}{8} \times r^{5} = 128 \to r^{5} = 128 \times 8 = 1024 \to r = 4.$
Thus, the series would be $\frac{1}{8}, \frac{1}{2}, 2, 8, 32, 128$ .
The third geometric mean would be 8.