Question: If $4, x_{1}, x_{2}, x_{3}, 28$ are in $\mathrm{AP}$, then $x_{3}=$ ?
(a) 19
(b) 23
(c) 22
(d) cannot be determined
Solution:
(c) 22
Here, a = 4, l = 28 and n = 5
Then, T5 = 28
⇒ a + (n - 1)d = 28
⇒ 4 + (5 - 1)d= 28
⇒ 4d = 24
⇒ d = 6
Hence, x3 = 28 - 6 = 22
