s7= 33, S12= 58. tentukan nilai a dan b dan s10?

u7=a+n-1.b
=a+6b
s7=n/2(2a+n-1.b)=33
=7/2(2a+6b)=33
=14a/2+42b/2=33
=7a+21b=33....per(1)
s12=12/2(2a+11b)=58
=12a+66b=58...pers (2)
eliminasi a dari pers 1&2
7a+21b=33|×12| 84a+252b=396
12a+66b=58|×7| 84a+462b=406
--------------------------- -
-210b=-10
b=-10/-210
b=1/21atau 0,04

subst b kepers1
7a+21b=33
7a+21(1/21)=33
7a+1=33
7a=33-1
7a=32
a=32/7

hit:s10=n/2(2a+n-1.b)
=10/2(2(32/7)+9(1/21)
= 5(64/7+9/21)
=5(9,14+0,42)
=5(9,56)
=47,8
smoga membantu jadikan jwbn terbaik ya

Kelas 9 Matematika
Bab 6 - Barisan dan Deret Bilangan

Deret Aritmatika

Sn = ½n [2a + (n - 1)b]

S7 = ½ × 7 [2a + (7 - 1)b]
33 = 3,5 (2a + 6b)
33 = 7a + 21b ...............(i)

S12 = ½ × 12 [2a + (12 - 1)b]
58 = 6 (2a + 11b)
58 = 12a + 66b ...........(ii)

7a + 21b = 33 ..............(×12)
12a + 66b = 58 ...........(×7)

Eliminasi nilai a
84a + 252b = 396
84a + 462b = 406
______________ _
-210b = -10
b = 1/21

Substitusikan nilai a
7a + 21b = 33
7a + 21(1/21) = 33
7a + 1 = 33
7a = 33 - 1
7a = 32
a = 32/7

S10 = ½ × 10 [2 × 32/7 + (10 - 1) 1/21]
S10 = 5 (64/7 + 9/21)
S10 = 5 (192/21 + 9/21)
S10 = 5 (201/21)
S10 = 1.005/21
S10 = 335/7