Jumlah bilangan-bilangan ganjil 3+5+7+...+k=440,maka k= ......

a.20
b.22
c.41
d.256
dengan caranya

Jumlah bilangan-bilangan ganjil 3+5+7+…+k=440,maka k= ……

Jawaban Terkonfirmasi

Dengan deret aritmatika:
a = 3
b = 2
Sn = 440
Maka,
$$begin{align}S_n&=440 \ frac{n}{2}(2a+(n-1)b)&=440 \ frac{n}{2}(2(3)+(n-1)2)&=440 \ frac{n}{2}(6+2n-2)&=440 \ frac{n}{2}(2n+4)&=440 \ n(n+2)&=440 \ n^2+2n&=440 \ n^2+2n-440&=0 \ (n+22)(n-20)&=0end{align}$
Karena n harus positif, maka:
n = 20
Sehingga,
k = a+(n-1)b
k = 3+(20-1)2
k = 3 + (19 x 2)
k = 3 + 38
k = 41

Jawaban Terkonfirmasi

A = 3
b = 2
$S_{n}$ = 440
$S_{n}$ = $frac{n}{2}(2a+(n-1)b)$
440 = $frac{n}{2}(6+(n-1)2)$
440 = $frac{n}{2}(6+2n-2)$
440 = $frac{n}{2}(4+2n)$
440 = $frac{4n+2n^{2} }{2}$
880 = $2n^{2}+4n$
0 =   $2n^{2}+4n$ -880
0 =   $n^{2}+2n$ -440
0   = (n+22)(n-20)
*n+22=0
n = -22 ( tidak mungkin karena min)
*n-20 = 0
n = 20