CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 23

Previous Year Board Questions

Question 1. The common difference of the A.P. \( \dfrac{1}{2b}, \dfrac{1 - 6b}{2b}, \dfrac{1 - 12b}{2b}, \dots \) is: (A) 2b (B) -2b (C) 3 (D) -3    [CBSE 2013, 1 Mark]
Answer:
\( d = \dfrac{1 - 6b}{2b} - \dfrac{1}{2b} = \dfrac{-6b}{2b} = -3 \). Answer: (D) -3.
In simple words: Subtract the first term from the second.

Teacher's Note:
1. The b cancels.

Question 2. How many three-digit natural numbers are divisible by 7? [CBSE 2013, 1 Mark]
Answer:
The numbers are 105, 112, ..., 994. \( 994 = 105 + (n - 1)7 \Rightarrow n = \) 128.
In simple words: Find the first and last three-digit multiples of 7.

Teacher's Note:
1. 105 = 7 x 15 and 994 = 7 x 142, so 142 - 15 + 1 = 128.

Question 3. The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term. [CBSE 2012, 4 Marks]
Answer:
\( a + 23d = 2(a + 9d) \Rightarrow a = 5d \). \( a_{72} = a + 71d = 76d \) and \( a_{15} = a + 14d = 19d \). So \( a_{72} = 4a_{15} \).
In simple words: Express a in terms of d.

Teacher's Note:
1. 76 = 4 x 19.

Question 4. The nth term of an A.P. is \( -4n + 15 \). Find the sum of the first 20 terms of this A.P. [CBSE 2013, 3 Marks]
Answer:
\( a_1 = 11 \), \( a_{20} = -65 \). \( S_{20} = \dfrac{20}{2}(11 - 65) = \) -540.
In simple words: Find the first and last terms.

Teacher's Note:
1. d = -4.

Question 5. The next term of \( \sqrt{7}, \sqrt{28}, \sqrt{63}, \dots \) is: (A) \( \sqrt{70} \) (B) \( \sqrt{84} \) (C) \( \sqrt{97} \) (D) \( \sqrt{112} \)    [CBSE 2013, 1 Mark]
Answer:
The terms are \( \sqrt{7}, 2\sqrt{7}, 3\sqrt{7} \), so the next is \( 4\sqrt{7} = \sqrt{112} \). Answer: (D).
In simple words: Simplify each surd.

Teacher's Note:
1. The A.P. has d = √7.

Question 6. The sum of the first n terms of an A.P. is \( 5n - n^2 \). Find the nth term of this A.P. [CBSE 2014, 2 Marks]
Answer:
\( a_n = S_n - S_{n-1} = 5n - n^2 - [5(n - 1) - (n - 1)^2] = 5 - (2n - 1) = \) 6 - 2n.
In simple words: a_n = S_n - S_(n-1).

Teacher's Note:
1. Check: a_1 = 4 = S_1.

Question 7. The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. [CBSE 2014, 3 Marks]
Answer:
\( S_7 = 63 \Rightarrow a + 3d = 9 \). \( S_{14} = 63 + 161 = 224 \Rightarrow 2a + 13d = 32 \). Solving, a = 3, d = 2. \( a_{28} = 3 + 27 \times 2 = \) 57.
In simple words: S_14 is the sum of the first 7 and the next 7 terms.

Teacher's Note:
1. Two equations in a and d.

Question 8. The 14th term of an A.P. is twice its 8th term. If its 6th term is -8, find the sum of its first 20 terms. [CBSE 2014, 3 Marks]
Answer:
\( a + 13d = 2(a + 7d) \Rightarrow a = -d \). \( a_6 = a + 5d = 4d = -8 \Rightarrow d = -2, a = 2 \). \( S_{20} = 10[4 + 19(-2)] = \) -340.
In simple words: Use the two conditions.

Teacher's Note:
1. a = -d.

Question 9. Find the 60th term of the A.P. 8, 10, 12, ..., if it has a total of 60 terms, and hence find the sum of its last 10 terms. [CBSE 2014, 4 Marks]
Answer:
\( a_{60} = 8 + 59 \times 2 = \) 126. The last 10 terms are \( a_{51}, \dots, a_{60} \) with \( a_{51} = 8 + 100 = 108 \). Sum \( = \dfrac{10}{2}(108 + 126) = \) 1170.
In simple words: The last ten terms are terms 51 to 60.

Teacher's Note:
1. They form an A.P. with d = 2.

Question 10. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term. [CBSE 2015, 2 Marks]
Answer:
\( a + 3d = 0 \Rightarrow a = -3d \). \( a_{25} = a + 24d = 21d \), \( a_{11} = a + 10d = 7d \). So \( a_{25} = 3a_{11} \).
In simple words: Express a in terms of d.

Teacher's Note:
1. 21 = 3 x 7.

Question 11. For what value of k will \( k + 9 \), \( 2k - 1 \) and \( 2k + 7 \) be the consecutive terms of an A.P.? [CBSE 2015, 1 Mark]
Answer:
\( 2(2k - 1) = (k + 9) + (2k + 7) \Rightarrow 4k - 2 = 3k + 16 \Rightarrow \) k = 18.
In simple words: Middle term = average of the others.

Teacher's Note:
1. Check: 27, 35, 43.

Question 12. If the ratio of the sum of the first n terms of two A.P.s is (7n + 1) : (4n + 27), find the ratio of their mth terms. [CBSE 2015, 3 Marks]
Answer:
The ratio of the mth terms equals the ratio of the sums for \( n = 2m - 1 \) (since \( a_m = \dfrac{S_{2m-1}}{2m - 1} \)). Hence \( \dfrac{7(2m - 1) + 1}{4(2m - 1) + 27} = \) \( \dfrac{14m - 6}{8m + 23} \).
In simple words: The mth term is the middle term of the first (2m - 1) terms.

Teacher's Note:
1. Use S_n = n x (middle term) when n is odd.

Question 13. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference. [CBSE 2016, 3 Marks]
Answer:
\( 400 = \dfrac{n}{2}(5 + 45) \Rightarrow n = 16 \). \( 45 = 5 + 15d \Rightarrow d = \dfrac{8}{3} \). n = 16, d = 8/3.
In simple words: Use S = n(a + l)/2 first.

Teacher's Note:
1. Then use l = a + (n - 1)d.

Question 14. What is the common difference of an A.P. in which \( a_{21} - a_7 = 84 \)? [CBSE 2017, 1 Mark]
Answer:
\( 14d = 84 \Rightarrow \) d = 6.
In simple words: The terms are 14 places apart.

Teacher's Note:
1. a_21 - a_7 = 14d.