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If K is the sum of the reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?
  • a)
    1/12
  • b)
    1/10
  • c)
    1/8
  • d)
    1/6
  • e)
    1/4
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If K is the sum of the reciprocals of the consecutive integers from 43...
Given that  Notice that 1/43 is the larges term and 1/48 is the smallest term.
If all 6 terms were equal to 1/43, then the sum would be 6/43=~1/7, but since actual sum is less than that, then we have that K<1/7.
If all 6 terms were equal to 1/48, then the sum would be 6/48=1/8, but since actual sum is more than that, then we have that K>1/8.
Therefore, 1/8<K<1/7. So, K must be closer to 1/8 than it is to 1/6.
Answer: C.
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Most Upvoted Answer
If K is the sum of the reciprocals of the consecutive integers from 43...
Understanding the Problem
To find K, we need to calculate the sum of the reciprocals of the integers from 43 to 48. This can be expressed as:
K = 1/43 + 1/44 + 1/45 + 1/46 + 1/47 + 1/48
Calculating Each Reciprocal
Let's calculate the approximate values of each reciprocal:
- 1/43 ≈ 0.0233
- 1/44 ≈ 0.0227
- 1/45 ≈ 0.0222
- 1/46 ≈ 0.0217
- 1/47 ≈ 0.0213
- 1/48 ≈ 0.0208
Summing the Values
Now, we sum these approximate values:
K ≈ 0.0233 + 0.0227 + 0.0222 + 0.0217 + 0.0213 + 0.0208
Calculating this gives:
K ≈ 0.131
Comparing with Options
Now, we compare K with the given options:
- 1/12 ≈ 0.0833
- 1/10 = 0.1
- 1/8 = 0.125
- 1/6 ≈ 0.1667
- 1/4 = 0.25
K ≈ 0.131 is closest to 0.125 (1/8).
Conclusion
Thus, the correct answer is option 'C', which corresponds to 1/8. This confirms that K falls nearest to this value among the options provided.
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If K is the sum of the reciprocals of the consecutive integers from 43 to 48, inclusive, then K is closest in value to which of the following?a)1/12b)1/10c)1/8d)1/6e)1/4Correct answer is option 'C'. Can you explain this answer?
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