Section A (1 Mark each)
Q1. (b) 6
$S_1 = 3(1)^2+1 = 4 = a_1$. $S_2 = 3(2)^2+2 = 14$. $a_2 = S_2 - S_1 = 10$. $d = 10-4=6$.
Q2. (a) 40
Reversing the AP: 49, 46, 43, 40... $a=49, d=-3$. $a_4 = 49 + 3(-3) = 40$.
Q3. (c) 0
$7a_7 = 11a_{11} \Rightarrow 7(a+6d) = 11(a+10d) \Rightarrow 7a+42d = 11a+110d \Rightarrow 4a+68d=0 \Rightarrow a+17d=0 \Rightarrow a_{18}=0$.
Q4. (b) 100
Difference between corresponding terms of two APs with same $d$ is constant. $a_n - b_n = a_1 - b_1$.
Q5. (a) Both A and R are true and R is the correct explanation of A.
19 is the 10th odd number ($2n-1=19 \Rightarrow n=10$). Sum = $10^2 = 100$.
Section B (2 Marks each)
Q6. $a=6, d=7, l=216$.
$216 = 6 + (n-1)7 \Rightarrow 210 = 7(n-1) \Rightarrow n-1=30 \Rightarrow n=31$.
Middle term is $\frac{31+1}{2} = 16$th term. $a_{16} = 6 + 15(7) = 6 + 105 = 111$. [2]
Q7. $a=3, d=12$. Let term be $a_n$.
$a_n = a_{54} + 132 \Rightarrow a + (n-1)d = a + 53d + 132$.
$(n-1)12 = 53(12) + 132 \Rightarrow n-1 = 53 + 11 \Rightarrow n = 65$. 65th term [2]
Q8. Numbers: 10, 13, ..., 97.
$97 = 10 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow n=30$.
Sum = $\frac{30}{2}(10 + 97) = 15(107) = 1605$. [2]
Section C (3 Marks each)
Q9. $S_7 = 49 \Rightarrow 7^2$. $S_{17} = 289 \Rightarrow 17^2$.
This implies $S_n = n^2$.
Alternatively, solve $2a+6d=14$ and $2a+16d=34$ to get $a=1, d=2$.
$S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}(2n) = n^2$. [3]
Q10. $S_7 = 700, n=7, d=-20$.
$700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 2a = 320 \Rightarrow
a = 160$.
Prizes: 160, 140, 120, 100, 80, 60, 40. [3]
Section D (Case Study 1 - 4 Marks)
(i) Distance = $2 \times 5 = 10$ m. [1]
(ii) Distance = $2 \times (5+3) = 16$ m. [1]
(iii) AP: 10, 16, 22... $n=10$.
$S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = 370$ m. [2]
Section E (Case Study 2 - 4 Marks)
(i) Total length = 2.5 m = 250 cm. Gap = 25 cm.
Number of rungs $n = \frac{250}{25} + 1 = 10 + 1 = 11$. [1]
(ii) $a=25, l=45$. Middle rung is average: $\frac{25+45}{2} = 35$ cm. [1]
(iii) Total length = $S_{11} = \frac{11}{2}(25 + 45) = \frac{11}{2}(70) = 11 \times 35 = 385$ cm. [2]