Arithmetic Progressions | Chapter Assessment | English | Grade 10

Which of the following is NOT an A.P? (i) a, a+1, a+2, a+3 … (ii) 4, 9, 14, 19 (iii) 4, 5, 4, 4, 5, 4, 4, 5 ...

• All

• (ii) and (iii)

• Only (iii)

• (i) and (iii)

Answer explanation

(i) a, a+1, a+2, a+3 … a₂ ─ a₁ = 1 a₃ ─ a₂ = 1 a₄ ─ a₃ = 1 So, this is an A.P. (ii) 4, 9, 14, 19 a₂ ─ a₁ = 5 a₃ ─ a₂ = 5 a₄ ─ a₃ = 5 So, this is an A. P. (iii) 4, 5, 4, 4, 5, 4, 4, 5…. a₂ ─ a₁ = 1 a₃ ─ a₂ = ─ 1 a₄ ─ a₃ = 0 Since a₃ ─ a₂ ≠ a₄ ─ a₃, this is not an A.P. Hence, the correct option is Option 3

Tags

CCSS.HSF.BF.A.2

Find the fourth term from the end of the A.P.: 2, 5, 8, …, 35.

• 26

• 12

• 29

• 23

Answer explanation

A.P. is 2, 5, 8, ..., 35 a = 2, d = 3 , l = 35 an = a + (n ─ 1) d 35 = 2 + (n ─ 1) 3 33 = 3n ─ 3 36 = 3n n = 12 So the A.P. has 12 terms. Fourth term from the end will be 9th term. a₉ = 2 + (9 ─ 1)3 a₉ = 2 + 24 = 26 So the correct answer is option1

Tags

CCSS.HSF.BF.A.2

Sum of the first 55 terms of an AP is 3300. What will be the 28th term of this AP?

• 60

• 120

• Cannot find the 28th term as the information given is insufficient

Answer explanation

Sum of 55 terms = 3300 n = 55 S₅₅ = n/2 (2a + (n ─ 1)d) 3300 = 55 / 2 ( 2a + 54d) 3300 = 55 / 2 x 2 (a + 27d) 3300 = 55 (a + 27 d) ………… (1) the 28th term = a + (28 ─ 1) d = a +27d From eq. 1, we get, 3300 / 55 = a +27d 60 = a +27d Hence, the 28th term is 60 So the correct answer is Option 1

Tags

CCSS.HSF.BF.A.2

Hari joined a company with a salary of Rs 15000 per month. He was given an increment of Rs 5000 after the first year, Rs.7000 after the second year and Rs. 9000 after the third year. Make a list of the monthly salary earned by Hari every year and interpret whether it is an AP or not.

• Not an AP

• It is a A.P. With common difference = Rs. 5000

• It is a A.P. With common difference = Rs. 2000

Answer explanation

Joining salary = Rs. 15000 Salary after one year = Rs 15000 + 5000 = Rs. 20000 Salary after two years = Rs 20000 + 7000 = Rs. 27000 Salary after three years = Rs 27000 + 9000 = Rs. 36000 List, 15000, 20000, 27000, 36000, …… Since 20000 - 15000 ≠ 27000 - 20000, this is not an AP So the correct option is Option 1 If Option 2 and Option 3 are chosen then these are incorrect. The student must have listed the increments Hari received every year instead of the salary he earned.

Tags

CCSS.HSF.BF.A.2

Ashish saves a part of his salary every month. In the first three months, he saves Rs. 100, Rs. 150, and Rs. 200 respectively. In which month will he save Rs. 1500?

• In 33rd month

• In 29th month

• In 27th month

Answer explanation

We can list Ashish's savings as 100, 150, 200, ... This is an AP with a = 100, d = 50 The nth term, aₙ = 1500 = a + (n ─ 1) d 1500 = 100 + (n ─ 1) 50 1400 = 50n ─ 50 50n = 1450 n = 1450 / 50 = 29 So Ashish will save 1500 Rs. in the 29th month. So the correct option is Option 2

Tags

CCSS.HSF.BF.A.2

Ramji borrowed Rs. 4000 and agreed to repay with a total interest of Rs. 500 in 10 instalments, each instalment being less than the previous one by Rs. 10. What will be the first and last instalment?

• First Instalment = Rs. 445 Last Instalment = Rs. 355

• First Instalment = Rs. 405 Last Instalment = Rs. 495

• First Instalment = Rs. 495 Last Instalment = Rs. 405

Answer explanation

Amount borrowed = Rs. 4000 Interest = Rs. 500 Total amount to be paid = Rs. 4500 n = 10 , d = ─ 10 ( as each instalment is less than the preceding one) S = n / 2 [ 2a + (n ─ 1) d ] 4500 = 5 [ 2a + 9 (─10)] 4500 = 5 (2a ─ 90) 4500 = 10 a ─ 450 4500 + 450 = 10 a 4950 = 10 a a = 495 Rs. The last instalment will be the tenth instalment l = a + (10 ─ 1) d l = 495 ─ 90 = 405 Rs. Hence, the correct answer is Option3 If Option 1 is chosen then it is incorrect. Here a and l have been calculated considering only Rs 4000 as sum to be repaid and not added the interest If Option 2 is chosen then also it is incorrect. Here the common difference d was taken as 10 and not ─ 10

Tags

CCSS.HSF.BF.A.2

In a children's balloon race, a bucket is placed at the starting point, which is 2m away from the first balloon and the other balloons are placed 3m apart in a straight line.There are 12 balloons in the line. A child starts from the bucket, picks up the nearest balloon, runs back with it, drops it in the bucket, runs back to pick up the next balloon and the again runs back to drop it in the bucket. The child continues the same way till all the balloons are in the bucket. Find the total distance run by the child.

• 222 m

• 444 m

• 666 m

• 888 m

Answer explanation

Distance of nearest balloon from the starting point = 2m Number of balloons = 12 The distance of the balloons are: 2, 5, 8, 11 ….. These numbers form an AP with a = 2 and d = 3 We know, S₁₂ = 12/2 [2 x 2 + (12 ─ 1)3] S₁₂ = 6 ( 4 + 11 x 3) = 6(37) = 222 As everytime the child has to run back to the bucket therefore the total distance that the child has to run will be 2 times of S₁₂ So,the total distance that the child runs = 2 x 222 m = 444 m So the correct option is Option 2

Tags

CCSS.HSF.BF.A.2