Solve:

Solution:

(i) 25 + 22 + 19 + 16 + ... + x = 115

Here, $a=25, d=-3, S_{n}=115$

We know:

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 115=\frac{n}{2}[2 \times 25+(n-1) \times(-3)]$

$\Rightarrow 115 \times 2=n[50-3 n+3]$

$\Rightarrow 230=n(53-3 n)$

$\Rightarrow 230=53 n-3 n^{2}$

$\Rightarrow 3 n^{2}-53 n+230=0$

By quadratic formula:

$n=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Substituting $a=3, b=-53$ and $c=230$, we get:

$n=\frac{53 \pm \sqrt{(53)^{2}-4 \times 3 \times 230}}{2 \times 3}=\frac{46}{6}, 10$

$\Rightarrow n=10$, as $n \neq \frac{46}{6}$

$\therefore a_{n}=a+(n-1) d$

$\Rightarrow x=25+(10-1)(-3)$

$\Rightarrow x=25-27=-2$

(ii) 1 + 4 + 7 + 10 + ... + x = 590

Here, a = 1, d = 3,

We know:

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 590=\frac{n}{2}[2 \times 1+(n-1) \times(3)]$

$\Rightarrow 590 \times 2=n[2+3 n-3]$

$\Rightarrow 3 n^{2}-n-1180=0$

By quadratic formula:

$n=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Substituting $a=3, b=-1$ and $c=-1180$, we get:

$\Rightarrow n=\frac{1 \pm \sqrt{(1)^{2}+4 \times 3 \times 1180}}{2 \times 3}=\frac{-118}{6}, 20$

$\Rightarrow n=20$, as $n \neq \frac{-118}{6}$

$\therefore a_{n}=x=a+(n-1) d$

$\Rightarrow x=1+(20-1)(3)$

$\Rightarrow x=1+60-3=58$