2

We have

b_{k} = 5 k a_{k} - 2 a_{k}

Let $b_{k}$ be a function of $a_{k}$ . For every k, we have

f (a_{k}) = 5 k a_{k} - 2 a_{k}

and

f^{'} (a_{k}) = \frac{5 k}{2 a _{k}} - 2

Let $f^{'} (x) = 0$ , we have

\frac{5 k}{2 a _{k}} = 2 ⟹ a_{k} = \frac{5 k}{4} ⟹ a_{k} = \frac{5 k}{16}

Substituting $a_{k} = \frac{5 k}{16}$ back into $f (a_{k})$

5 k \cdot \frac{5 k}{16} - 2 \cdot \frac{5 k}{16} = \frac{5 k}{8}

The derivative function is strictly decreasing, so we can conclude that

max (5 k a_{k} - 2 a_{k}) = \frac{5 k}{8}

Add all numbers together we have

k = 1 \sum n b_{k} \leq k = 1 \sum n \frac{5 k}{8} = \frac{5}{8} \frac{n ( n + 1 )}{2} = \frac{5 n ( n + 1 )}{16} = 2025

So we have

n (n + 1) = \frac{2025 \times 16}{5} = 6480

We know that $6480 = 80 \times 81$ , so we get

n = 80

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