Q1. Reverse AP: $253, 248, \dots, 3. Common difference $d = -5. $a_{20} = a + 19d = 253 + 19(-5) = 253 - 95 = 158$.
Q2. $S_n = 4n - n^2. $a_1 = S_1 = 4(1)-1 = 3. $S_{10} = 40-100 = -60, $S_9 = 36-81 = -45. $a_{10} = S_{10} - S_9 = -60 - (-45) = -15$.
Q3. AP: $105, 112, \dots, 994. $a=105, d=7, a_n=994 \rightarrow 994 = 105 + (n-1)7 \rightarrow n=128. $S_{128} = 64(105 + 994) = 70336$.
Q4. $a_4 + a_8 = 24 \rightarrow 2a + 10d = 24. $a_6 + a_{10} = 44 \rightarrow 2a + 14d = 44. Solving gives $d = 5, a = -13. First three terms: $-13, -8, -3$.
Q5. $S_n = 78 \rightarrow \frac{n}{2}[2(24) + (n-1)(-3)] = 78 \rightarrow 3n^2 - 51n + 156 = 0 \rightarrow n = 4, 13. Sum of 4 terms and 13 terms are both 78 because terms after 4th term cancel each other out.
