A) \[\frac{{{k}^{2}}}{72}\]
B) \[\frac{72}{k}\]
C) \[\frac{72}{n}\]
D) none of these
Correct Answer: C
Solution :
| \[\frac{1}{{{d}_{1}}}+\frac{1}{{{d}_{2}}}+\frac{1}{{{d}_{3}}}+.....+\frac{1}{{{d}_{k}}}=\frac{1}{n}\left[ \frac{n}{{{d}_{1}}}+\frac{n}{{{d}_{2}}}+\frac{n}{{{d}_{3}}}+.......+\frac{n}{{{d}_{5}}} \right]\] |
| Now \[\frac{n}{{{d}_{1}}},\frac{n}{{{d}_{2}}},.....\]will also be divisor of the number, i.e., |
| \[\frac{n}{{{d}_{j}}}={{d}_{i}}\]for the same j and i. \[\Rightarrow \] \[\frac{1}{{{d}_{1}}}+\frac{1}{{{d}_{2}}}+......\frac{1}{{{d}_{k}}}=\frac{1}{n}[{{d}_{1}}+{{d}_{2}}+....]=\frac{72}{n}\] |
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