Check whether – 150 is a term of the AP

Solution:

$a=11, d=-3$

Let if possible $\mathrm{t}_{\mathrm{n}}=-150$

$\Rightarrow a+(n-1) d=-150$

$\Rightarrow 11+(n-1) \times(-3)=-150$

$\Rightarrow 11-3 n+3=-150$

$\Rightarrow 14-3 n=-150$

$\Rightarrow 3 n=14+150=164$

$\Rightarrow \mathrm{n}=\frac{164}{3}=54 \frac{2}{3}$

It is not possible because n is to be natural number.

Hence, – 150 cannot be a term of the AP.