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

Given $p = \sum_{j=1}^{n} \log_{10} 2^j and$q = \sum_{j=1}^{n} \log_{10} 5^j$.<br><br>Simplifying$p$and$q$:<br><br>$p = \sum_{j=1}^{n} j \log_{10} 2 = \log_{10} 2 \sum_{j=1}^{n} j = \dfrac{n(n+1)}{2} \log_{10} 2$<br><br>$q = \sum_{j=1}^{n} j \log_{10} 5 = \log_{10} 5 \sum_{j=1}^{n} j = \dfrac{n(n+1)}{2} \log_{10} 5$<br><br>Adding$p$and$q$:<br><br>$p + q = \dfrac{n(n+1)}{2} (\log_{10} 2 + \log_{10} 5) = \dfrac{n(n+1)}{2} \log_{10} (2 \times 5) = \dfrac{n(n+1)}{2} \log_{10} 10$<br><br>Since$\log_{10} 10 = 1$, we get:<br><br>$p + q = \dfrac{n(n+1)}{2}$<br><br>Given$p + q = 15$, we have:<br><br>$\dfrac{n(n+1)}{2} = 15$<br><br>Now, finding$q - p$:<br><br>$q - p = \dfrac{n(n+1)}{2} \log_{10} 5 - \dfrac{n(n+1)}{2} \log_{10} 2$<br><br>$q - p = \dfrac{n(n+1)}{2} (\log_{10} 5 - \log_{10} 2)$<br><br>$q - p = 15 \log_{10} $\left(\dfrac{5}{2}\right)<br><br>$q - p = 15 \log_{10} 2.5