Solutions: Arithmetic Progressions - Test 2

Q1. (b) 400

Sum of first $n$ odd numbers is $n^2$. Here $n=20$, so $20^2 = 400$.

Q2. (c) 10

$S_1 = 3(1)^2 + 1 = 4 \Rightarrow a_1 = 4$.
$S_2 = 3(2)^2 + 2 = 12 + 2 = 14$.
$a_2 = S_2 - S_1 = 14 - 4 = 10$.

Q3. (a) 275

AP: 5, 10, ..., 50. $S_{10} = \frac{10}{2}(5 + 50) = 5(55) = 275$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Using $S_n = \frac{n}{2}(a+l)$ for $1, 2, ..., n$ gives $\frac{n(1+n)}{2}$.

Q5. $a=2, d=5, n=10$.
$S_{10} = \frac{10}{2}[2(2) + (10-1)5] = 5[4 + 45] = 5(49) = 245$. [2]

Q6. $S_{1000} = \frac{1000(1000+1)}{2} = 500(1001) = 500500$. [2]

Q7. Multiples of 8: 8, 16, ...
$a=8, d=8, n=15$.
$S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15}{2}(128) = 15(64) = 960$. [2]

Q8. $a=24, d=-3, S_n=78$.
$78 = \frac{n}{2}[2(24) + (n-1)(-3)] \Rightarrow 156 = n[48 - 3n + 3] \Rightarrow 156 = n[51 - 3n]$.
$156 = 51n - 3n^2 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0$.
$(n-4)(n-13) = 0 \Rightarrow n=4$ or $n=13$. Both are valid. [3]

Q9. Odd numbers: 1, 3, 5, ..., 49.
$a=1, l=49$. Number of terms $n = 25$.
$S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2}(50) = 25 \times 25 = 625$. [3]

Q10. $a=17, l=350, d=9$.
$l = a + (n-1)d \Rightarrow 350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n=38$. [2]
$S_{38} = \frac{38}{2}(17 + 350) = 19(367) = 6973$. [3]

(i) Penalty forms an AP: 200, 250, 300...
$a=200, d=50$. $a_{10} = 200 + 9(50) = 200 + 450 = 650$. ₹ 650 [1]

(ii) $S_{30} = \frac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] = 15(1850) = 27750$. ₹ 27750 [2]

(iii) Difference is simply the common difference $d$. ₹ 50 [1]