In an A.P, T₁= 22, T_n = —11, S_n = 66, find n.

We know that, T_n = a + (n – 1)d

So for T_1,

We have T₁=a + (1 – 1)d = a

According to the question, a = 22

Now, we have T_n = – 11

Ie., a + (n – 1)d = – 11

So, 22 + (n – 1)d = – 11

⇒ (n – 1)d = – 22 – 11

⇒ (n – 1)d = – 33

Now we have that, S_n = × (2a + (n – 1)d)

From the question we can say that, S_n = 66

So, we have,

⇒ 66= × (2a + (n – 1)d)

⇒ 132 = n × (2a + (n – 1)d)

We have (n – 1)d = – 33 and a = 22, so we put that in the above equation.

⇒ 132 = n × (2(22) – 33)

⇒ 132 = n × (44 – 33)

⇒ 132= n × 11

⇒ n =

⇒ n = 12

∴ value of n is 12.