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1. 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 15 for the first km and 8 for each additional km.

(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.

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

(iv) The amount of money in the account every year, when 10000 is deposited at compound interest at 8 % per annum.

Solution :(i) Taxi fare after each km

Fare = ₹15 for the first km and ₹8 for each additional km

Fare after each km:

  • 1 km → 15
  • 2 km → 15 + 8 = 23
  • 3 km → 23 + 8 = 31
  • 4 km → 31 + 8 = 39

So, the list is: 15, 23, 31, 39, …

Difference between successive terms:

  • 23 − 15 = 8
  • 31 − 23 = 8
  • 39 − 31 = 8

✔ This forms an A.P. because the common difference is 8.

(ii) Amount of air in a cylinder

Each time, 1/4 of the remaining air is removed.

Let the initial air be 1 unit:

  • After first removal → 34\frac{3}{4}
  • After second removal → 34×34=916\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}
  • After third removal → 2764\frac{27}{64}

The sequence is: 1, 34\frac{3}{4}​, 916\frac{9}{16}​, 2764\frac{27}{64}……….

Here, the differences are not constant.

❌ This does not form an A.P. because the decrease is not by a fixed amount.

(iii) Cost of digging a well

Cost = ₹150 for first metre, increases by ₹50 for each metre.

Costs are:

  • 1st metre → 150
  • 2nd metre → 200
  • 3rd metre → 250
  • 4th metre → 300

Sequence: 150, 200, 250, 300, …

Difference between terms:

  • 200 − 150 = 50
  • 250 − 200 = 50
  • 300 − 250 = 50

✔ This forms an A.P. with common difference 50.

(iv) Money deposited at compound interest

₹10,000 deposited at 8% compound interest per year.

Amount each year:

  • 1st year → 10000
  • 2nd year → 10800
  • 3rd year → 11664
  • 4th year → 12597.12

The differences are not constant.

❌ This does not form an A.P. because the increase is by a fixed percentage, not a fixed amount.

Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) a = 10, d = 10 (ii) a = –2, d = 0 (iii) a = 4, d = – 3 (iv) a = – 1, d = 1 2 (v) a = – 1.25, d = – 0.25

For an arithmetic progression (A.P.), the terms are:
a, a + d, a + 2d, a + 3d

(i) a = 10, d = 10

First four terms:

  • 1st term = 10
  • 2nd term = 10 + 10 = 20
  • 3rd term = 10 + 20 = 30
  • 4th term = 10 + 30 = 40

✅ A.P.: 10, 20, 30, 40

(ii) a = –2, d = 0

First four terms:

  • 1st term = –2
  • 2nd term = –2 + 0 = –2
  • 3rd term = –2 + 0 = –2
  • 4th term = –2 + 0 = –2

✅ A.P.: –2, –2, –2, –2
(This is a constant A.P.)

(iii) a = 4, d = –3

First four terms:

  • 1st term = 4
  • 2nd term = 4 – 3 = 1
  • 3rd term = 1 – 3 = –2
  • 4th term = –2 – 3 = –5

✅ A.P.: 4, 1, –2, –5

(iv) a = –1, d = 1/2

First four terms:

  • 1st term = –1
  • 2nd term = –1 + 1/2 = –1/2
  • 3rd term = –1/2 + 1/2 = 0
  • 4th term = 0 + 1/2 = 1/2

✅ A.P.: –1, –1/2, 0, 1/2

(v) a = –1.25, d = –0.25

First four terms:

  • 1st term = –1.25
  • 2nd term = –1.25 – 0.25 = –1.50
  • 3rd term = –1.50 – 0.25 = –1.75
  • 4th term = –1.75 – 0.25 = –2.00

✅ A.P.: –1.25, –1.50, –1.75, –2.00

For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .

(ii) – 5, – 1, 3, 7, . . .

iii)1/3, 5/3, 9/3, 13/3, …

(iv) 0.6, 1.7, 2.8, 3.9, . . .

(i) 3, 1, –1, –3, …

  • First term, a = 3
  • Common difference,
    d=1−3=−2d = 1 – 3 = -2

✅ Answer:
a = 3, d = –2

(ii) –5, –1, 3, 7, …

  • First term, a = –5
  • Common difference,
    d=−1−(−5)=4d = -1 – (-5) = 4

✅ Answer:
a = –5, d = 4

(iii) 1/3, 5/3, 9/3, 13/3, …

  • First term, a = 1/3
  • Common difference,
    d=53−13=43d = \frac{5}{3} – \frac{1}{3} = \frac{4}{3}

✅ Answer:
a = 1/3, d = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, …

  • First term, a = 0.6
  • Common difference,
    d=1.7−0.6=1.1d = 1.7 – 0.6 = 1.1d=1.7−0.6=1.1

✅ Answer:
a = 0.6, d = 1.1

  1. Which of the following are APs ? If they form an AP, find the common difference d and
    write three more terms.
    (i) 2, 4, 8, 16, . . . .(ii) 5 7 2, , 3, ,
    2 2 . . . (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . (iv) – 10, – 6, – 2, 2, . . .
    (vi) 0.2, 0.22, 0.222, 0.2222, . . .
    (vii) 0, – 4, – 8, –12, . . .

(i) 2, 4, 8, 16, …

Differences:
4 − 2 = 2, 8 − 4 = 4, 16 − 8 = 8 (not equal)

❌ Not an A.P.

(ii) 52,72,32,…\frac{5}{2}, \frac{7}{2}, \frac{3}{2}, …25​,27​,23​,…

Differences:
72−52=1\frac{7}{2} – \frac{5}{2} = 127​−25​=1
32−72=−2\frac{3}{2} – \frac{7}{2} = -223​−27​=−2

❌ Not an A.P.

(iii) –1.2, –3.2, –5.2, –7.2, …

Differences:
–3.2 − (–1.2) = –2
–5.2 − (–3.2) = –2
–7.2 − (–5.2) = –2

✅ This is an A.P.

  • Common difference (d) = –2

Next three terms:
–9.2, –11.2, –13.2

(iv) –10, –6, –2, 2, …

Differences:
–6 − (–10) = 4
–2 − (–6) = 4
2 − (–2) = 4

✅ This is an A.P.

  • Common difference (d) = 4

Next three terms:
6, 10, 14

(vi) 0.2, 0.22, 0.222, 0.2222, …

Differences are not equal

❌ Not an A.P.

(vii) 0, –4, –8, –12, …

Differences:
–4 − 0 = –4
–8 − (–4) = –4
–12 − (–8) = –4

✅ This is an A.P.

  • Common difference (d) = –4

Next three terms:
–16, –20, –24

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