1. **Problem:** If $p-1$, $p+1$, and $2p+3$ are in arithmetic progression (A.P.), find the value of $p$. **Formula:** In an A.P., the middle term is the average of the first and third terms: $$2 \times (p+1) = (p-1) + (2p+3)$$ **Solution:** $$2p + 2 = p - 1 + 2p + 3$$ $$2p + 2 = 3p + 2$$ $$2p + 2 - 3p - 2 = 0$$ $$-p = 0 \Rightarrow p = 0$$ 2. **Problem:** In a group of 20 people, 5 can't swim. Find the probability that a randomly selected person can swim. **Formula:** Probability = (Number of favorable outcomes) / (Total outcomes) **Solution:** Number who can swim = $20 - 5 = 15$ $$P = \frac{15}{20} = \frac{3}{4}$$ 3. **Problem:** Determine the nature of lines represented by equations $2x = 3y + 6$ and $15y = 6x - 18$. **Rewrite equations:** $$2x - 3y = 6$$ $$6x - 15y = 18$$ **Check ratios:** $$\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-3}{-15} = \frac{1}{5}, \quad \frac{c_1}{c_2} = \frac{6}{18} = \frac{1}{3}$$ Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, lines are intersecting. 4. **Problem:** A bag contains 5 red balls and $n$ green balls. Probability of drawing a green ball is three times that of a red ball. Find $n$. **Formula:** $$P(\text{green}) = 3 \times P(\text{red})$$ **Solution:** $$\frac{n}{5 + n} = 3 \times \frac{5}{5 + n}$$ $$n = 15$$ 5. **Problem:** Sum of first $n$ terms of an A.P. is $3n^2 + n$. Common difference $d=6$. Find the first term $a$. **Formula:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ **Given:** $$S_n = 3n^2 + n$$ **Equate:** $$3n^2 + n = \frac{n}{2} [2a + (n-1)6]$$ $$6n + 2 = 2a + 6n - 6$$ $$2a = 8 \Rightarrow a = 4$$ 6. **Problem:** Probability of drawing a face card from a deck of 52 cards. **Number of face cards:** 3 face cards per suit $\times$ 4 suits = 12 $$P = \frac{12}{52} = \frac{3}{13}$$ 7. **Problem:** Probability of getting "at most one head" when a fair coin is tossed twice. **Sample space:** HH, HT, TH, TT **At most one head:** 0 or 1 head = TT, HT, TH (3 outcomes) $$P = \frac{3}{4}$$ 8. **Problem:** Number of terms in an A.P. with first term $a_1 = -14$, fifth term $a_5 = 2$, last term $a_n = 62$. **Formula:** $$a_n = a_1 + (n-1)d$$ **Find common difference $d$:** $$a_5 = a_1 + 4d = 2$$ $$-14 + 4d = 2 \Rightarrow 4d = 16 \Rightarrow d = 4$$ **Find $n$:** $$62 = -14 + (n-1)4$$ $$(n-1)4 = 76 \Rightarrow n-1 = 19 \Rightarrow n = 20$$ 9. **Problem:** For system $2x + 3y = 7$ and $ax + (a+b)y = 28$ to have infinite solutions, find $a$ and $b$. **Condition for infinite solutions:** $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$ $$\frac{2}{a} = \frac{3}{a+b} = \frac{7}{28} = \frac{1}{4}$$ From first equality: $$\frac{2}{a} = \frac{1}{4} \Rightarrow a = 8$$ From second equality: $$\frac{3}{a+b} = \frac{1}{4} \Rightarrow a + b = 12$$ Substitute $a=8$: $$8 + b = 12 \Rightarrow b = 4$$ 10. **Problem:** Solve system: $$217x + 131y = 913$$ $$131x + 217y = 827$$ **Use Cramer's rule:** $$D = \begin{vmatrix}217 & 131 \\ 131 & 217\end{vmatrix} = 217 \times 217 - 131 \times 131 = 47089 - 17161 = 29928$$ $$D_x = \begin{vmatrix}913 & 131 \\ 827 & 217\end{vmatrix} = 913 \times 217 - 131 \times 827 = 198121 - 108337 = 89784$$ $$D_y = \begin{vmatrix}217 & 913 \\ 131 & 827\end{vmatrix} = 217 \times 827 - 913 \times 131 = 179459 - 119603 = 59856$$ $$x = \frac{D_x}{D} = \frac{89784}{29928} = 3$$ $$y = \frac{D_y}{D} = \frac{59856}{29928} = 2$$ **Final answers:** 1) $p=0$ 2) $\frac{3}{4}$ 3) Intersecting 4) $n=15$ 5) $a=4$ 6) $\frac{3}{13}$ 7) $\frac{3}{4}$ 8) $20$ 9) $a=8$, $b=4$ 10) $x=3$, $y=2$