Suku ke 5 suatu deret aritmatika = 40 dan suku ke 8 = 45. Tentukan suku pertama dan beda serta hitunglah jumlulah 10 suku pertama

U5 = a + 4b = 40
U8 = a + 7b = 45
--------------------------(-)
- 3b = - 5
b = 5/3

a + 4b = 40
a + 4(5/3) = 40
a + 20/3 = 40
a = 40 - 20/3
a = 120/3 - 20/3
a = 100/3

U10 = a + 9b
= 100/3 + 9(5/3)
= 100/3 + 45/3
= 145/3

S10 = ½ . 10 ( a + U10)
= 5(100/3 + 145/3)
= 5 × 245/3
= 1225/3
= 408 ⅓