A man starts repaying a loan as the first instalment of 10000.

To Find: what amount will he pay in the 30th instalment.

Given: first instalment =10000 and it increases the instalment by 500 every month.

$\therefore$ So it form an AP with first term is 10000 , common difference 500 and number of instalment is 30

Formula Used: $T_{n}=a+(n-1) d$

(Where $\mathrm{a}$ is first term, $\mathrm{T}_{\mathrm{n}}$ is nth term and $\mathrm{d}$ is common difference of given AP)

$\therefore \mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \Rightarrow \mathrm{T}_{\mathrm{n}}=10000+(30-1) 500 \Rightarrow \mathrm{T}_{\mathrm{n}}=10000+29 \times 500$

$\therefore \mathrm{T}_{\mathrm{n}}=10000+14500 \Rightarrow \mathrm{T}_{\mathrm{n}}=24,500$

So, he will pay 24,500 in the 30th instalment.