diketahui 3+5+7+...+k=440 tentukan nilai k?

3 + 5 + 7 + ... + k = 440 => a = 3, b = 2
Sn = 440
n/2 (2a + (n - 1)b) = 440
n/2 (2(3) + (n - 1).2) = 440
n/2 (6 + 2n - 2) = 440
n/2 (2n + 4) = 440
n² + 2n = 440
n² + 2n - 440 = 0
(n + 22)(n - 20) = 0
n = -22 atau n = 20

k = U20
= a + (20 - 1)b
= 3 + 19(2)
= 3 + 38
= 41

perhatikan bahwa deret di atas adalah deret aritmatika
[tex]b=u_{2}-u_{1}[/tex]
[tex]=5-3[/tex]
[tex]=2[/tex]

[tex]u_{n}=a+(n-1)b[/tex]
[tex]k=3+2(n-1)[/tex]
[tex]k=2n+1[/tex]

[tex]S_{n}=\frac{n}{2}(u_{1}+u_{n})[/tex]
[tex]440=\frac{n}{2}(3+k)[/tex]
[tex]880=n(3+2n+1)[/tex]
[tex]880=2n^{2}+4n[/tex]
[tex]n^{2}+2n-440=0[/tex]
[tex](n+22)(n-22)=0[/tex]
[tex]n=20[/tex]
[tex]k=2n+1=41[/tex]