Concept: 0.5 Mark
Application: 0.5 Mark
Let the first term of the AP be a, and the common difference be d.
Given: Sum of 99 terms is 198
S99=198⇒992[2a+(99−1)d]=198 [Sn=n2[2a+(n−1)d]
⇒2a+98d=198×299=4
⇒a+49d=2
⇒a+(50−1)d=2
⇒a50=2 [nth term of an AP=a+(n−1)d]
∴ The 50th term is 2.