Respuesta :
so they start off with 5 bricks, then they add 2 bricks on the left side and 2 bricks on the right side, namely 4 bricks, so the first row is 5 bricks, the next row is 5+4 or 9 bricks and so on.
5, 9, 13, 17.... <--- as you can see the "common difference" is 4.
[tex]\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=5\\ d=4\\ n=25 \end{cases} \\\\\\ a_{25}=5+(25-1)(4)\implies a_{25}=5+(24)(4) \\\\\\ a_{25}=5+96\implies a_{25}=101\\\\ -------------------------------[/tex]
[tex]\bf \textit{ sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ ----------\\ a_1=5\\ a_{25}=101\\ n=25 \end{cases} \\\\\\ S_{25}=\cfrac{25(5+101)}{2}\implies S_{25}=\cfrac{25(106)}{2}\implies S_{25}=1325[/tex]
5, 9, 13, 17.... <--- as you can see the "common difference" is 4.
[tex]\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=5\\ d=4\\ n=25 \end{cases} \\\\\\ a_{25}=5+(25-1)(4)\implies a_{25}=5+(24)(4) \\\\\\ a_{25}=5+96\implies a_{25}=101\\\\ -------------------------------[/tex]
[tex]\bf \textit{ sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ ----------\\ a_1=5\\ a_{25}=101\\ n=25 \end{cases} \\\\\\ S_{25}=\cfrac{25(5+101)}{2}\implies S_{25}=\cfrac{25(106)}{2}\implies S_{25}=1325[/tex]
The answer would be a. 1325 bricks.
COmmon difference= 4Use arithmetic sequence:An=A1_(n-1)dA25=5_(25-1)(4)A25=5(24)(4)A25=5+96A25=101
Sn=n(A1+An)/2S25=25(5+101)/2=25(106)/2=1325
COmmon difference= 4Use arithmetic sequence:An=A1_(n-1)dA25=5_(25-1)(4)A25=5(24)(4)A25=5+96A25=101
Sn=n(A1+An)/2S25=25(5+101)/2=25(106)/2=1325