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NDA 2 2025 Maths, question 24

Topic: Algebra

Passage · Q23 to Q24

Let p=∑j=1nlog⁡102jp=\sum _{j=1}^{n}\log _{10}2^{j} and q=∑j=1nlog⁡105jq=\sum _{j=1}^{n}\log _{10}5^{j}.

Q24

If p+q=15p+q=15, then what is q−pq-p equal to?

  1. (a) log⁡102.5\log _{10}2.5
  2. (b) 5log⁡102.55\log _{10}2.5
  3. (c) 10log⁡102.510\log _{10}2.5
  4. (d) 15log⁡102.515\log _{10}2.5Correct answer

Solution.

Answer: (d) 15log⁡102.515\log _{10}2.5

  1. Similarly, q−p=Σj log₁₀(5/2)=[n(n+1)/2]log₁₀2.5.
  2. Since p+q=n(n+1)/2=15, the difference is 15 log₁₀2.5.