A) A.P.
B) G.P.
C) H.P.
D) None of the above
Correct Answer: C
Solution :
If the numbers are \[\frac{1}{x},\ \frac{1}{y},\ \frac{1}{z}\], then \[x={{\log }_{2}}3,\ y={{\log }_{2}}2\ .\ 3=1+{{\log }_{2}}3\] and\[z=2+{{\log }_{2}}3\]. Therefore \[=\frac{1.({{10}^{91}}-1)}{10-1}=\frac{{{({{10}^{13}})}^{7}}-1}{{{10}^{13}}-1}\times \frac{{{10}^{13}}-1}{10-1}\] are in A.P. Hence\[\frac{1}{x},\ \frac{1}{y},\ \frac{1}{z}\ \ \ i.e.\], the given numbers are in H.P.
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