Answer:
d. 5490
Step-by-step explanation:
[tex]\begin{aligned}\displaystyle \sf \sum_{n=1}^{60} 3n & =3(1)+3(2)+3(3)+...+3(49)+3(50)\\ & = 3+6+9+...+147+150\end{aligned}[/tex]
Therefore, this is an arithmetic series with:
- first term (a) = 3
- common difference (d) = 3
Sum of the first n terms of an arithmetic series:
[tex]S_n=\dfrac12n[2a+(n-1)d][/tex]
Therefore, the sum of the first 60 terms:
[tex]\begin{aligned}S_{60} &=\dfrac12(60)[2(3)+(60-1)3]\\ & =30[6+(59)3]\\ & = 30[6+177]\\ & = 30[183]\\ & = 5490 \end{aligned}[/tex]
