The first term of an AP is 5,

Solution:

$a=5$, last term $t_{n}=45$ and $S_{n}=400$

$\mathrm{S}_{\mathrm{n}}=400 \Rightarrow \frac{\mathrm{n}}{2}\left\{\mathrm{t}_{1}+\mathrm{t}_{\mathrm{n}}\right\}=400$

$\Rightarrow \frac{n}{2}\{5+45\}=400 \Rightarrow \frac{n}{2} \times 50=400$

$\Rightarrow \mathrm{n}=16$

Now, $\mathrm{t}_{\mathrm{n}}=45 \quad \Rightarrow \mathrm{t}_{16}=45$

$\Rightarrow \mathrm{a}+15 \mathrm{~d}=45 \quad \Rightarrow 5+15 \mathrm{~d}=45$

$\Rightarrow 15 \mathrm{~d}=40 \quad \Rightarrow \mathrm{d}=8 / 3$