Question

Jumlah deret aritmatika 2 + 5 + 8 +....+ k= 345, maka k = ?

Answer 1

Verified answer

K=3n-1
Sn=n/2.(a+k)
Sn=n/2.(2+3n-1)
345=n/2.(3n+1)
690=3n²+n
3n²+n-690=0
(n-15)(3n+46)=0
n-15=0
n=15

k=3(15)-1 =44

Answer 2

Mapel : Matematika
Bab : Barisan dan Deret

Deret Aritmatika

2 + 5 + 8 + ... + k = 345

Diketahui :
a = 2
b = 5 - 2 = 3
Un = k
Sn = 345

Ditanyakan :
Nilai k = ... ?

Penyelesaian :
       Sn = n/2 (2a + (n-1) b )
      345 = n/2 (2.2 + (n-1) 3 )
345 x 2 = n ( 4 + 3n - 3 )
     690  = n ( 1 + 3n )
     690  = 3n² + n
3n² + n - 690 = 0
(n - 15) (3n + 46) = 0

n - 15 = 0
      n  = 0 + 15
      n  = 15

Un = bn + a - b
   k = 3(15) + 2 - 3
   k = 45 - 1
   k = 44

Jadi, nilai k adalah 44