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Question 1: The 10th term of the AP: 2, 7, 12, ... is:
Solution: (A) 47
Step 1: Here $a=2, d=7-2=5$.
Step 2: $a_{10} = a + (10-1)d = 2 + 9(5) = 2 + 45 = 47$. -
Question 2: Which term of the AP: 21, 18, 15, ... is -81?
Solution: (B) 35
Step 1: $a=21, d=-3, a_n=-81$.
Step 2: $-81 = 21 + (n-1)(-3) \Rightarrow -102 = -3(n-1) \Rightarrow 34 = n-1 \Rightarrow n=35$. -
Question 3: The sum of the first five multiples of 3 is:
Solution: (A) 45
Step 1: AP is 3, 6, 9, 12, 15.
Step 2: Sum $= 3+6+9+12+15 = 45$. -
Question 4: If $k, 2k-1$ and $2k+1$ are three consecutive terms of an AP, the value of $k$ is:
Solution: (B) 3
Step 1: For AP, $2b = a+c$. So, $2(2k-1) = k + (2k+1)$.
Step 2: $4k - 2 = 3k + 1 \Rightarrow k = 3$. -
Question 5: The common difference of the AP $\frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p}, \dots$ is:
Solution: (C) -1
Step 1: $d = a_2 - a_1 = \frac{1-p}{p} - \frac{1}{p} = \frac{1-p-1}{p} = \frac{-p}{p} = -1$. -
Question 6: If the $n^{th}$ term of an AP is $3n + 4$, then the common difference is:
Solution: (B) 3
Step 1: $a_1 = 3(1)+4 = 7$, $a_2 = 3(2)+4 = 10$.
Step 2: $d = 10 - 7 = 3$. (Coefficient of $n$ is always $d$). -
Question 7: The sum of first $n$ terms of an AP is $3n^2 + n$. Then the second term of this AP is:
Solution: (B) 10
Step 1: $S_1 = a_1 = 3(1)^2 + 1 = 4$.
Step 2: $S_2 = a_1 + a_2 = 3(2)^2 + 2 = 14$.
Step 3: $a_2 = S_2 - S_1 = 14 - 4 = 10$. -
Question 8: The sum of first 20 odd natural numbers is:
Solution: (C) 400
Step 1: Sum of first $n$ odd numbers is $n^2$.
Step 2: Here $n=20$, so Sum $= 20^2 = 400$. -
Question 9: If the 7th term of an AP is 4 and its common difference is -4, then its first term is:
Solution: (D) 28
Step 1: $a_7 = a + 6d = 4$.
Step 2: $a + 6(-4) = 4 \Rightarrow a - 24 = 4 \Rightarrow a = 28$. -
Question 10: The 4th term from the end of the AP: -11, -8, ..., 49 is:
Solution: (B) 40
Step 1: Reverse the AP: 49, 46, ... ($d = -3$).
Step 2: 4th term $= a + 3d = 49 + 3(-3) = 49 - 9 = 40$.