Deret arit S4=17 dan S8=58, U1

s4 = 17
sn = n/2 (2a + (n-1) b)
s4 = 2(2a + 3b)
17 = 4a + 6b

s8 = 58
s8 = 4 ( 2a + 7b)
58 = 8a + 28b

4a + 6b = 17. (×2)
8a + 28b = 58. (×1)

8a + 12b = 34
8a + 28b = 58. (kurangkan)

-16b = -24
b = 24/16
b = 3/2

4a + 6b = 17
4a + 6(3/2) = 17
4a + 9 = 17
4a = 8
a = 2

S4 = 17
4/2 (2a + (4-1)b) = 17
2 (2a + 3b) = 17
4a + 6b = 17

S8 = 58
8/2 (2a + (8-1)b) = 58
4 (2a + 7b) = 58
2 (2a + 7b) = 29
4a + 14b = 29

eliminasi
4a + 14b = 29
4a + 6b = 17
→ 8b = 12
b = 3/2

4a + 6b = 17
4a + 6(3/2) = 17
4a + 9 = 17
4a = 8
a = 2

semoga bermanfaat