Answer 1

Given that, d = 5, S₉ = 75

`S_n = n/2[2a+(n-1)d]`

`S_9 = 9/2[2a+(9-1)5]`

75=9/2(2a+40)

25 = 3(a + 20)

25 = 3a + 60

3a = 25 − 60

`a = (-35)/3`

a_n = a + (n − 1)d

a₉ = a + (9 − 1) (5)

`= (-35)/3+8(5)`

`=(-35+120)/3=85/3`