Arithmetic Progressions l Exercise 5.1 l Class 10 l NCERT Solutions

Exercise 5.1 – Arithmetic Progressions (Stepwise Solutions)

EXERCISE 5.1 – Stepwise Solutions


1. Check whether the following form an AP

(i) Taxi fare: 15, 23, 31, 39, …

Step 1: Find differences.

23 – 15 = 8

31 – 23 = 8

39 – 31 = 8

Since difference is constant, it forms an AP.

(ii) Air removed each time:

Remaining air: 1, 3/4, 9/16, …

Differences are not equal.

Not an AP.

(iii) Cost of digging: 150, 200, 250, 300, …

Common difference = 50

Forms an AP.

(iv) Compound interest: 10000, 10800, 11664, …

Differences are not equal.

Not an AP.


2. Write first four terms

(i) a = 10, d = 10

10, 20, 30, 40

(ii) a = –2, d = 0

–2, –2, –2, –2

(iii) a = 4, d = –3

4, 1, –2, –5

(iv) a = –1, d = 1/2

–1, –1/2, 0, 1/2

(v) a = –1.25, d = –0.25

–1.25, –1.5, –1.75, –2


3. Find first term and common difference

(i) 3, 1, –1, –3

a = 3

d = 1 – 3 = –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

d = 1.1


4. Which are APs?

(i) 2, 4, 8, 16

Differences not equal → Not AP

(ii) 2, 5/2, 3, 7/2

d = 1/2 → AP

Next terms: 4, 9/2, 5

(iii) –1.2, –3.2, –5.2, –7.2

d = –2 → AP

Next: –9.2, –11.2, –13.2

(iv) –10, –6, –2, 2

d = 4 → AP

Next: 6, 10, 14

(v) 3, 3+√2, 3+2√2, 3+3√2

d = √2 → AP

Next: 3+4√2, 3+5√2, 3+6√2

(vi) 0.2, 0.22, 0.222

Differences not equal → Not AP

(vii) 0, –4, –8, –12

d = –4 → AP

Next: –16, –20, –24

(viii) –1/2, –1/2, –1/2

d = 0 → AP

(ix) 1, 3, 9, 27

Not AP

(x) a, 2a, 3a, 4a

d = a → AP

(xi) a, a², a³

Not AP

(xii) √2, √8, √18, √32

= √2, 2√2, 3√2, 4√2

d = √2 → AP

(xiii) √3, √6, √9, √12

Not AP

(xiv) 1², 3², 5², 7²

1, 9, 25, 49 → Not AP

(xv) 1², 5², 7², 73

Not AP

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