Question

Pada deret aritmatika diketahui bahwa u1+u6=16 dan u2+u5+u8=42. Nilai u7 adalah....

Answer 1

Un = a + (n - 1)b

maka :

u1 = a + (1 - 1)b = a
u6 = a + (6 - 1)b = a + 5b

a + a + 5b = 16
2a + 5b = 16

u2 = a + (2 - 1)b = a + b
u5 = a + (5 - 1)b = a + 4b
u8 = a + (8 - 1)b = a + 7b

a + b + a + 4b + a + 7b = 42
3a + 12b = 42

3a + 12b = 42 | x2
2a + 5b = 16    | x3

6a + 24b = 84
6a + 15b = 48 -
9b = 36
b = 36/9
b = 4

2a + 5b = 16
2a + 5(4) = 16
2a + 20 = 16
2a = 16 - 20
2a = -4
a = -4/2
a = -2

u7 = a + (7 - 1)b
= a + 6b
= -2 + 6(4)
= -2 + 24
= 22

Answer 2

U1+U6=16
a+a+5b = 16
2a + 5b = 16 ............................... (1)

U2+U5+U8=42
a+b+a+4b+a+7b=42
3a+12b = 42
a+4b = 14 ....................................(2)

lalu substitusikan keduanya :

2a+5b = 16                    
2a+8b= 28
        b = 4
        a = -2
U7 = a + 6b
     = -2 + 6(4)
     = 22