+36915 a1 = 3 d = 5
xn = a1 + d(n-1)
253= 3 + 5(n-1)
50 = n-1
n= 51
Sum = S = (n/2) × (2a + (n−1)d) = 6528
+26367 Find the sum of the numbers \(3,8,13,\ldots,253\)
arithmetic series:
\(\mathbf{n=\ ?}\)
\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)\cdot d \quad &| \quad a_n = 253,\ a_1=3,\ d=5 \\\\ 253 &=& 3+(n-1)\cdot 5 \quad &| \quad -3 \\ 250 &=& (n-1)\cdot 5 \quad &| \quad :5 \\ 50 &=&n-1 \quad &| \quad +1 \\ 51 &=&n\\ \mathbf{ n } &=& \mathbf{51} \\ \hline \end{array}\)
sum:
\(\begin{array}{|rcll|} \hline sum &=& \left(\dfrac{a_1+a_n}{2} \right) \cdot n \quad &| \quad a_n = 253,\ a_1=3,\ n=51 \\\\ sum &=& \left(\dfrac{3+253}{2} \right) \cdot 51 \\ sum &=& \left(\dfrac{256}{2} \right) \cdot 51 \\ sum &=& 128 \cdot 51 \\ \mathbf{sum} &=& \mathbf{6528} \\ \hline \end{array}\)
