Question

In an A.P. (with usual notations) : given d = 5, S₉ = 75, find a and a₉ 

Answer 1

d = 5, S₉ = 75
a_n = a + (n – 1)d
a₉ = a + (9 – 1) x 5
= a + 40                          ...(i)
S₉ = `n/(2)[2a + (n - 1)d]`

75 = `(9)/(2)[2a + 8 xx 5]`

`(150)/(9)` = 2a + 40

2a = `(150)/(9) - 40`

= `(50)/(3) - 40`

2a = `(-70)/(3)`

⇒ a = `(-70)/(2 xx 3)`

a = `(-35)/(3)`
From (i),
a₉ = a + 40
= `(-35)/(3) + 40`

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

= `(85)/(3)`

∴ a = `(-35)/(3), a_9 = (85)/(3)`.