The first term of an AP is 2. The sum of 8 terms of AP is 156. Find the common difference of AP. Given that the sum of n term of AP 116, Find n.

Badilisha maadili uliyopewa: $S_8 = 156$, $n = 8$, na $a = 2$.

$$156 = \frac{8}{2}(2(2) + (8-1)d)$$

$$156 = 4(4 + 7d)$$ $$156 = 16 + 28d$$ $$156 - 16 = 28d$$ $$140 = 28d$$ $$d = \frac{140}{28}$$ $$d = 5$$

Tofauti ya kawaida ni $\boxed{5}$.

a) ii)
Tunapewa kwamba jumla ya maneno $n$ ya AP ni $S_n = 116$.
Kutoka sehemu (i), tunajua $a = 2$ na $d = 5$.
Tunahitaji kupata $n$.

$S_n = \frac{n}{2}(2a + (n-1)d)$.
Badilisha maadili yanayojulikana: $S_n = 116$, $a = 2$, na $d = 5$.

$$116 = \frac{n}{2}(2(2) + (n-1)5)$$

$$116 = \frac{n}{2}(4 + 5n - 5)$$ $$116 = \frac{n}{2}(5n - 1)$$

Panua pande zote mbili kwa 2:

$$232 = n(5n - 1)$$ $$232 = 5n^2 - n$$

Hapa, $a = 5$, $b = -1$, $c = -232$.

$$n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(5)(-232)}}{2(5)}$$ $$n = \frac{1 \pm \sqrt{1 + 4640}}{10}$$ $$n = \frac{1 \pm \sqrt{4641}}{10}$$

Kwa kuwa $n$ lazima iwe nambari kamili chanya kwa idadi ya maneno katika AP, na $\sqrt{4641}$ si nambari kamili, hakuna thamani kamili ya $n$ inayokidhi masharti haya.

$$n \approx \frac{1 \pm 68.12}{10}$$ $$n \approx \frac{1 + 68.12}{10} \approx 6.912 \quad \text{au} \quad n \approx \frac{1 - 68.12}{10} \approx -6.712$$

Kwa kuwa $n$ lazima iwe nambari kamili chanya, hakuna suluhisho kamili la $n$ kwa jumla ya 116.

Hakuna thamani kamili chanya ya $n$ kwa jumla ya 116.