Answer:
[tex]S_{40}=6560[/tex]
Step-by-step explanation:
Given the sequence
[tex]4,12,20,28,\cdots[/tex]
We know that:
The first term, a=4
Also, 21-4=20-12=28-20=8
Therefore, the sequence is an arithmetic sequence with:
Common difference, d=8
For an arithmetic sequence, the sum
[tex]S_n=\dfrac{n}{2}[2a+(n-1)d] \\$Therefore$:\\\\S_{40}=\dfrac{40}{2}[2(4)+(40-1)*8] \\=20(8+39*8)\\=20(8+312)\\=20*320\\=6560[/tex]
The sum of the first 40 terms is 6560.