Which term of the AP :

Solution:

$a=3, d=12$

Let us suppose $t_{n}=t_{54}+132$

$\Rightarrow a+(n-1) d=a+53 d+132$

$\Rightarrow(\mathrm{n}-1) \mathrm{d}-53 \mathrm{~d}=132$

$\Rightarrow\{n-1-53) d=132$

$\Rightarrow(n-54) \times 12=132$

$\Rightarrow \mathrm{n}-54=11$

$\Rightarrow \mathrm{n}=65$

Hence, $\mathrm{t}_{65}$ is 132 more than $\mathrm{t}_{54}$.