Section A (1 Mark each)
Q1. (a) 32
$a=5, d=3$. $a_{10} = a + 9d = 5 + 9(3) = 5 + 27 = 32$.
Q2. (b) 3
For AP: $2(2k-1) = k + (2k+1) \Rightarrow 4k-2 = 3k+1 \Rightarrow k=3$.
Q3. (c) $n^2$
Sum of first $n$ odd numbers is $n^2$.
Q4. (a) Both A and R are true and R is the correct explanation of A.
Ratios are $4/2=2, 8/4=2$. It is a GP, not AP. Reason defines AP correctly.
Section B (2 Marks each)
Q5. $a=21, d=-3, a_n=-81$.
$-81 = 21 + (n-1)(-3) \Rightarrow -102 = -3(n-1) \Rightarrow 34 = n-1 \Rightarrow n=35$.
35th term [2]
Q6. $a_3=5 \Rightarrow a+2d=5$. $a_7=9 \Rightarrow a+6d=9$.
Subtracting: $4d=4 \Rightarrow d=1$.
$a + 2(1) = 5 \Rightarrow a=3$.
AP: 3, 4, 5, 6... [2]
Q7. Numbers: 12, 15, ..., 99.
$a=12, d=3, a_n=99$.
$99 = 12 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow 29 = n-1 \Rightarrow n=30$.
30 numbers [2]
Section C (3 Marks each)
Q8. $a=8, d=-5, n=22$.
$S_{22} = \frac{22}{2}[2(8) + (22-1)(-5)] = 11[16 + 21(-5)]$.
$= 11[16 - 105] = 11(-89) = -979$. [3]
Q9. $a_4+a_8=24 \Rightarrow a+3d+a+7d=24 \Rightarrow 2a+10d=24 \Rightarrow
a+5d=12$.
$a_6+a_{10}=44 \Rightarrow a+5d+a+9d=44 \Rightarrow 2a+14d=44 \Rightarrow a+7d=22$.
Subtracting: $2d=10 \Rightarrow d=5$.
$a + 25 = 12 \Rightarrow a = -13$.
AP: -13, -8, -3 [3]
Section D (5 Marks)
Q10. $S_n = 4n - n^2$.
(i) First term $a_1 = S_1 = 4(1) - 1^2 = 3$. [1]
(ii) Sum of first two terms $S_2 = 4(2) - 2^2 = 8 - 4 = 4$. [1]
(iii) Second term $a_2 = S_2 - S_1 = 4 - 3 = 1$. [1]
(iv) $n$th term $a_n = S_n - S_{n-1} = (4n - n^2) - [4(n-1) - (n-1)^2]$.
$= 4n - n^2 - [4n - 4 - (n^2 - 2n + 1)] = 4n - n^2 - [4n - 4 - n^2 + 2n - 1]$.
$= 4n - n^2 - [-n^2 + 6n - 5] = 4n - n^2 + n^2 - 6n + 5 = 5 - 2n$. [2]
Section E (Case Study - 4 Marks)
(i) $a_3 = 600, a_7 = 700$.
$a+2d=600, a+6d=700 \Rightarrow 4d=100 \Rightarrow d=25$.
$a + 50 = 600 \Rightarrow a = 550$. Production in 1st year = 550. [2]
(ii) $a_{10} = a + 9d = 550 + 9(25) = 550 + 225 = 775$. [1]
(iii) $S_7 = \frac{7}{2}[2(550) + 6(25)] = \frac{7}{2}[1100 + 150] = \frac{7}{2}[1250] = 4375$. [1]