Find X in the following series. 1, 512, 8, 343, 27, 216, 64, 125, X

121

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144

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125

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134

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Solution

The correct option is C 125
The given series is a combination of two separate arithmetic progressions (alternate terms)
Series I = 1, 8, 27, 64, X
Series II = 512, 343, 216, 125
X = 125

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Similar questions

Find the cube-roots of:

(i) $\frac{27}{64}$ (ii) $\frac{125}{216}$ (iiii) $\frac{343}{512}$

(iv) $64 \times 729$ (v) $64 \times 27$ (vi) $729 \times 8000$

(vii) $3375 \times 512$

Find the cube-roots of:

(i) -216 (ii) -512 (iiii) -1331

(iv) $- \frac{27}{125}$ (v) $\frac{- 64}{343}$ (vi) $- \frac{512}{343}$

(vii) -2197 (viii) -5832 (ix) -2744000

(8 - 27), (125 - 216), (343 - 512), (1009 - 1331)

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