(Q24).In an AP:

(vi) Given an = 4, d = 2, Sn = -14, find n and a.

Given an = a + (n – 1) d = 4 ……. (1)

d = ; Sn = –

From (1); a + (n – 1) = 4

a = 4 – n +

a = – n

Given a = , d = , Sn =

Sn =
n
2
 [a + an]
- =
n
2
 [(-n)+] [∵ a = 6 – 2n]

- × = n ( – n)

⇒ n – n2 = –

⇒ n2 – n – = 0

⇒ n2 – n – = 0

⇒ n2 – n + n – = 0

⇒ (n – ) + (n – ) = 0

⇒ (n – ) (n + ) = 0

⇒ n = (or) n = –

∴ n =

Now a = 6 – 2n = 6 – 2 ×

= 6 – = -

∴ a = – ; n =