${log}_{3} 2, {log}_{6} 2$ and ${log}_{12} 2$ are in.

A . P .

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G . P .

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H . P .

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None of these

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Solution

The correct option is C $H . P .$
Given sequence is, ${log}_{3} 2, {log}_{6} 2$ and ${log}_{12} 2$
Reciprocal of this is , ${log}_{2} 3, {log}_{2} 6$ and ${log}_{2} 12$
Clearly $2 {log}_{2} 6 = {log}_{2} 3 + {log}_{2} 12$
Thus the later sequence is in an A.P
Therefore, given sequence is a H.P
Hence, option 'C' is correct.

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Similar questions

{log}_{3} 2, {log}_{6} 2, {log}_{12} 2

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log32,log62 and log122 are in.

The correct option is C H.P.Given sequence is, log32,log62 and log122Reciprocal of this is , log23,log26 and log212Clearly 2log26=log23+log212Thus the later sequence is in an A.PTherefore, given sequence is a H.P Hence, option 'C' is correct.

${log}_{3} 2, {log}_{6} 2$ and ${log}_{12} 2$ are in.