Q173 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Let $p = \sum_{j=1}^{n} \log_{10} 2^j$ and $q = \sum_{j=1}^{n} \log_{10} 5^j$.
If $p + q = 66$, then which one of the following is correct?
Let $p = \sum_{j=1}^{n} \log_{10} 2^j$ and $q = \sum_{j=1}^{n} \log_{10} 5^j$.
If $p + q = 66$, then which one of the following is correct?
- A$n < 7$
- B$7 < n < 9$
- Correct answerC$9 < n < 12$
- D$n > 12$
Solution
Given $p = \sum_{j=1}^{n} \log_{10} 2^j$ and $q = \sum_{j=1}^{n} \log_{10} 5^j$
Adding $p$ and $q$:
$p + q = \sum_{j=1}^{n} (\log_{10} 2^j + \log_{10} 5^j)$
$p + q = \sum_{j=1}^{n} \log_{10} (2^j \cdot 5^j)$
$p + q = \sum_{j=1}^{n} \log_{10} (10^j)$
$p + q = \sum_{j=1}^{n} j \log_{10} 10$
Since $\log_{10} 10 = 1$, we get:
$p + q = \sum_{j=1}^{n} j = \dfrac{n(n+1)}{2}$
Given $p + q = 66$:
$\dfrac{n(n+1)}{2} = 66$
$n(n+1) = 132$
$n^2 + n - 132 = 0$
$(n - 11)(n + 12) = 0$
Since $n$ must be a positive integer, $n = 11$.
The value $n = 11$ satisfies the condition $9 < n < 12$.
Answer: $9 < n < 12$
Adding $p$ and $q$:
$p + q = \sum_{j=1}^{n} (\log_{10} 2^j + \log_{10} 5^j)$
$p + q = \sum_{j=1}^{n} \log_{10} (2^j \cdot 5^j)$
$p + q = \sum_{j=1}^{n} \log_{10} (10^j)$
$p + q = \sum_{j=1}^{n} j \log_{10} 10$
Since $\log_{10} 10 = 1$, we get:
$p + q = \sum_{j=1}^{n} j = \dfrac{n(n+1)}{2}$
Given $p + q = 66$:
$\dfrac{n(n+1)}{2} = 66$
$n(n+1) = 132$
$n^2 + n - 132 = 0$
$(n - 11)(n + 12) = 0$
Since $n$ must be a positive integer, $n = 11$.
The value $n = 11$ satisfies the condition $9 < n < 12$.
Answer: $9 < n < 12$
Hence, the correct answer is option: 'C'.