exercise 5.1 solution class 10 answer

MATHEMATICS CLASS 10 – CHAPTER 5
ARITHMETIC PROGRESSIONS (EXERCISE 5.1 SOLUTIONS)

1. Q1. In which of the following situations does the list of numbers involved make an arithmetic progression and why?

   (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

   Solution:
   Taxi fare for 1 km = 15
   Taxi fare for first 2 kms = 15 + 8 = 23
   Taxi fare for first 3 kms = 23 + 8 = 31
   Taxi fare for first 4 kms = 31 + 8 = 39
   
   Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

   (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

   Solution:
   Let the volume of air initially be V litres.
   After every stroke, 3/4 of the air remains.
   Therefore, volumes will be V, 3V/4, (3V/4)², (3V/4)³ …
   
   There is no common difference between consecutive terms.
   Hence, it is not an A.P.

   (iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

   Solution:
   Cost for 1^st metre = 150
   Cost for 2^nd metre = 150 + 50 = 200
   Cost for 3^rd metre = 200 + 50 = 250
   Cost for 4^th metre = 250 + 50 = 300
   
   Thus, 150, 200, 250, 300 … forms an A.P. with common difference 50.

   (iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound interest at 8% per annum.

   Solution:
   Amount each year = 10000(1 + 8/100), 10000(1 + 8/100)², 10000(1 + 8/100)³ …
   
   No common difference exists.
   Therefore, this is not an A.P.

2. Q2. Write the first four terms of the A.P. when the first term a, and the common difference d are given as follows:

   (i) a = 10, d = 10
   Series: 10, 20, 30, 40, 50 …
   First four terms: 10, 20, 30, 40

   (ii) a = –2, d = 0
   Series: –2, –2, –2, –2 …
   First four terms: –2, –2, –2, –2

   (iii) a = 4, d = –3
   Series: 4, 1, –2, –5 …
   First four terms: 4, 1, –2, –5

   (iv) a = –1, d = 1/2
   Series: –1, –1/2, 0, 1/2 …
   First four terms: –1, –1/2, 0, 1/2

   (v) a = –1.25, d = –0.25
   Series: –1.25, –1.50, –1.75, –2.00 …
   First four terms: –1.25, –1.50, –1.75, –2.00

3. Q3. For the following A.P.s, write the first term and the common difference.

   (i) 3, 1, –1, –3 …
   a = 3, d = –2

   (ii) –5, –1, 3, 7 …
   a = –5, d = 4

   (iii) 1/3, 5/3, 9/3, 13/3 …
   a = 1/3, d = 4/3

   (iv) 0.6, 1.7, 2.8, 3.9 …
   a = 0.6, d = 1.1

4. Q4. Which of the following are A.Ps? If they form an A.P., find the common difference d and write three more terms.

   (i) 2, 4, 8, 16 … → Not an A.P.

   (ii) 2, 5/2, 3, 7/2 … → d = 1/2; Next terms: 4, 9/2, 5

   (iii) –1.2, –3.2, –5.2, –7.2 … → d = –2; Next terms: –9.2, –11.2, –13.2

   (iv) –10, –6, –2, 2 … → d = 4; Next terms: 6, 10, 14

   (v) 3, 3+√2, 3+2√2, 3+3√2 → d = √2; Next terms: 3+4√2, 3+5√2, 3+6√2

   (vi) 0.2, 0.22, 0.222, 0.2222 … → Not an A.P.

   (vii) 0, –4, –8, –12 … → d = –4; Next terms: –16, –20, –24

   (viii) –1/2, –1/2, –1/2, –1/2 … → d = 0; Next terms: –1/2, –1/2, –1/2

   (ix) 1, 3, 9, 27 … → Not an A.P.

   (x) a, 2a, 3a, 4a … → d = a; Next terms: 5a, 6a, 7a

   (xi) a, a², a³, a⁴ … → Not an A.P.

   (xii) √2, √8, √18, √32 … → d = √2; Next terms: √50, √72, √98

   (xiii) √3, √6, √9, √12 … → Not an A.P.

   (xiv) 1², 3², 5², 7² → Not an A.P.

   (xv) 1², 5², 7², 8² → d = 24; Next terms: 97, 121, 145