SSC CGL 4 Sep 2016 Paper

nth term of an arithmetic progression :
${\mathit{a}}_{\mathit{n}}$ = a + (n – 1) d
∴ ${\mathit{a}}_7$ = a + (7 – 1) d = a + 6d
${\mathit{a}}_{11}$ = a + (11 – 1) d = a + 10d
According to the question,
7 ${\mathit{a}}_7$ = 11 ${\mathit{a}}_{11}$
⇒ 7 (a + 6d) = 11 (a + 10d)
⇒ 7a + 42d = 11a + 110 d
⇒ 11a – 7a = 42d – 110d
⇒ 4a = – 68d
⇒ a = – 17d .... (i)
∴ ${\mathit{a}}_{18}$ = a + (18 – 1)d = a + 17d
= –17d + 17d
= 0