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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2025 · Mathematics · Basics of Mathematics

For the following two (02) items: Let p = ∑_ j=1 ^ n _ 10 2^j and q...

For the following two (02) items:
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.

If p+q=66p + q = 66, then which one of the following is correct?

Options

  1. An<7n < 7
  2. B7<n<97 < n < 9
  3. C9<n<129 < n < 12
  4. Dn>12n > 12

Correct answer

C. 9<n<129 < n < 12

Step-by-step solution

Given 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

Adding pp and qq:

p+q=∑j=1n(log⁡102j+log⁡105j)p + q = \sum_{j=1}^{n} (\log_{10} 2^j + \log_{10} 5^j)

p+q=∑j=1nlog⁡10(2j⋅5j)p + q = \sum_{j=1}^{n} \log_{10} (2^j \cdot 5^j)

p+q=∑j=1nlog⁡10(10j)p + q = \sum_{j=1}^{n} \log_{10} (10^j)

p+q=∑j=1njlog⁡1010p + q = \sum_{j=1}^{n} j \log_{10} 10

Since log⁡1010=1\log_{10} 10 = 1, we get:

p+q=∑j=1nj=n(n+1)2p + q = \sum_{j=1}^{n} j = \dfrac{n(n+1)}{2}

Given p+q=66p + q = 66:

n(n+1)2=66\dfrac{n(n+1)}{2} = 66

n(n+1)=132n(n+1) = 132

n2+n−132=0n^2 + n - 132 = 0

(n−11)(n+12)=0(n - 11)(n + 12) = 0

Since nn must be a positive integer, n=11n = 11.

The value n=11n = 11 satisfies the condition 9<n<129 < n < 12.

Answer: 9<n<129 < n < 12
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