Answer :
You need 16 more terms not 17 because you already have 4 terms in the series so if you add those 4 terms to the other 16 terms to the series you get a total of 20 terms in total. I hope this helped.
[tex]670 = n/2[2*5+(n-1)*3][/tex]
[tex]670 = n/2[10+(n-1)*3][/tex]
[tex]1340 = n[10+3n-3][/tex]
[tex]1340 = n[7+3n][/tex]
[tex]1340 = 7n+3n^{2}[/tex]
[tex]3n^{2}+7n-1340 = 0[/tex]
[tex]n^{2}+\frac{7}{3}n-\frac{1340}{3} = 0[/tex]
[tex]n^{2}+\frac{7}{3}n = \frac{1340}{3}[/tex]
[tex]n^{2}+\frac{7}{3}n+\frac{49}{36}[/tex] [tex]=[/tex] [tex]\frac{1340}{3}+\frac{49}{36}[/tex]
[tex]n^{2}+\frac{7}{3}n+\frac{49}{36} = \frac{16129}{36}[/tex]
[tex](n+\frac{7}{6})^2 = \frac{16129}{36}[/tex]
[tex]n+\frac{7}{6} = \sqrt{\frac{16129}{36}}[/tex]
[tex]n = -\frac{7}{6}+\sqrt{\frac{16129}{36}}[/tex]
[tex]n = -1.167 + 21.167[/tex]
[tex]n = 20[/tex]
[tex]670 = n/2[10+(n-1)*3][/tex]
[tex]1340 = n[10+3n-3][/tex]
[tex]1340 = n[7+3n][/tex]
[tex]1340 = 7n+3n^{2}[/tex]
[tex]3n^{2}+7n-1340 = 0[/tex]
[tex]n^{2}+\frac{7}{3}n-\frac{1340}{3} = 0[/tex]
[tex]n^{2}+\frac{7}{3}n = \frac{1340}{3}[/tex]
[tex]n^{2}+\frac{7}{3}n+\frac{49}{36}[/tex] [tex]=[/tex] [tex]\frac{1340}{3}+\frac{49}{36}[/tex]
[tex]n^{2}+\frac{7}{3}n+\frac{49}{36} = \frac{16129}{36}[/tex]
[tex](n+\frac{7}{6})^2 = \frac{16129}{36}[/tex]
[tex]n+\frac{7}{6} = \sqrt{\frac{16129}{36}}[/tex]
[tex]n = -\frac{7}{6}+\sqrt{\frac{16129}{36}}[/tex]
[tex]n = -1.167 + 21.167[/tex]
[tex]n = 20[/tex]
