Which term of the AP 53, 48, 43,

Given AP is 53, 48, 43, …

Whose, first term (a) = 53 and common difference (d) = 48 – 53 = – 5

Let nth term of the AP be the first negative term.

i.e., $\quad T_{n}^{-}<0 \quad\left[\because n\right.$th termof an AP, $\left.T_{n}=a+(n-1) d\right]$

$(a+(n-1) d)<0$

$\Rightarrow \quad 53+(n-1)(-5)<0$

$\Rightarrow \quad 53-5 n+5<0$

$\Rightarrow \quad 58-5 n<0 \Rightarrow 5 n>58$

$\Rightarrow \quad n>11.6 \Rightarrow n=12$

i.e., 12 th term is the first negative term of the given AP.

$T_{12}=a+(12-1) d=53+11(-5)$

$=53-55=-2<0$