Arithmetic Progressions | Exercise 5.1

Question 4

  1. Which of the following are APs ? If they form an AP, find the common difference dd and write three more terms.

(i) 2,4,8,16,…2, 4, 8, 16, \dots

(ii) 2,52,3,72,…2, \frac{5}{2}, 3, \frac{7}{2}, \dots

(iii) −1.2,−3.2,−5.2,−7.2,…-1.2, -3.2, -5.2, -7.2, \dots

(iv) −10,−6,−2,2,…-10, -6, -2, 2, \dots

(v) 3,3+2,3+22,3+32,…3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots

(vi) 0.2,0.22,0.222,0.2222,…0.2, 0.22, 0.222, 0.2222, \dots

(vii) 0,−4,−8,−12,…0, -4, -8, -12, \dots

(viii) −12,−12,−12,−12,…-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots

(ix) 1,3,9,27,…1, 3, 9, 27, \dots

(x) a,2a,3a,4a,…a, 2a, 3a, 4a, \dots

(xi) a,a2,a3,a4,…a, a^2, a^3, a^4, \dots

(xii) 2,8,18,32,…\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots

(xiii) 3,6,9,12,…\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots

(xiv) 12,32,52,72,…1^2, 3^2, 5^2, 7^2, \dots

(xv) 12,52,72,73,…1^2, 5^2, 7^2, 73, \dots

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Solution

We check if the difference between consecutive terms is constant. If it is, the sequence is an Arithmetic Progression (AP).

Step 1 — Check sequence (i)

Let's find the differences between terms.

a2−a1=4−2a_2 - a_1 = 4 - 2

=2= 2

a3−a2=8−4a_3 - a_2 = 8 - 4

=4= 4

The differences are not the same.

Not an AP\boxed{\text{Not an AP}}

Step 2 — Check sequence (ii)

Let's find the differences between terms.

a2−a1=52−2a_2 - a_1 = \frac{5}{2} - 2

=52−42= \frac{5}{2} - \frac{4}{2}

=12= \frac{1}{2}

a3−a2=3−52a_3 - a_2 = 3 - \frac{5}{2}

=62−52= \frac{6}{2} - \frac{5}{2}

=12= \frac{1}{2}

a4−a3=72−3a_4 - a_3 = \frac{7}{2} - 3

=72−62= \frac{7}{2} - \frac{6}{2}

=12= \frac{1}{2}

The common difference is constant.

d=12\boxed{d = \frac{1}{2}}

Let's find the next three terms.

a5=a4+d=72+12a_5 = a_4 + d = \frac{7}{2} + \frac{1}{2}

=82=4= \frac{8}{2} = 4

a6=a5+d=4+12a_6 = a_5 + d = 4 + \frac{1}{2}

=92= \frac{9}{2}

a7=a6+d=92+12a_7 = a_6 + d = \frac{9}{2} + \frac{1}{2}

=102=5= \frac{10}{2} = 5

Next three terms: 4,92,5\boxed{\text{Next three terms: } 4, \frac{9}{2}, 5}

Step 3 — Check sequence (iii)

Let's find the differences between terms.

a2−a1=−3.2−(−1.2)a_2 - a_1 = -3.2 - (-1.2)

=−3.2+1.2= -3.2 + 1.2

=−2.0= -2.0

a3−a2=−5.2−(−3.2)a_3 - a_2 = -5.2 - (-3.2)

=−5.2+3.2= -5.2 + 3.2

=−2.0= -2.0

a4−a3=−7.2−(−5.2)a_4 - a_3 = -7.2 - (-5.2)

=−7.2+5.2= -7.2 + 5.2

=−2.0= -2.0

The common difference is constant.

d=−2.0\boxed{d = -2.0}

Let's find the next three terms.

a5=a4+d=−7.2+(−2.0)a_5 = a_4 + d = -7.2 + (-2.0)

=−9.2= -9.2

a6=a5+d=−9.2+(−2.0)a_6 = a_5 + d = -9.2 + (-2.0)

=−11.2= -11.2

a7=a6+d=−11.2+(−2.0)a_7 = a_6 + d = -11.2 + (-2.0)

=−13.2= -13.2

Next three terms: −9.2,−11.2,−13.2\boxed{\text{Next three terms: } -9.2, -11.2, -13.2}

Step 4 — Check sequence (iv)

Let's find the differences between terms.

a2−a1=−6−(−10)a_2 - a_1 = -6 - (-10)

=−6+10= -6 + 10

=4= 4

a3−a2=−2−(−6)a_3 - a_2 = -2 - (-6)

=−2+6= -2 + 6

=4= 4

a4−a3=2−(−2)a_4 - a_3 = 2 - (-2)

=2+2= 2 + 2

=4= 4

The common difference is constant.

d=4\boxed{d = 4}

Let's find the next three terms.

a5=a4+d=2+4a_5 = a_4 + d = 2 + 4

=6= 6

a6=a5+d=6+4a_6 = a_5 + d = 6 + 4

=10= 10

a7=a6+d=10+4a_7 = a_6 + d = 10 + 4

=14= 14

Next three terms: 6,10,14\boxed{\text{Next three terms: } 6, 10, 14}

Step 5 — Check sequence (v)

Let's find the differences between terms.

a2−a1=(3+2)−3a_2 - a_1 = (3 + \sqrt{2}) - 3

=2= \sqrt{2}

a3−a2=(3+22)−(3+2)a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2})

=3+22−3−2= 3 + 2\sqrt{2} - 3 - \sqrt{2}

=2= \sqrt{2}

a4−a3=(3+32)−(3+22)a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2})

=3+32−3−22= 3 + 3\sqrt{2} - 3 - 2\sqrt{2}

=2= \sqrt{2}

The common difference is constant.

d=2\boxed{d = \sqrt{2}}

Let's find the next three terms.

a5=a4+d=(3+32)+2a_5 = a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2}

=3+42= 3 + 4\sqrt{2}

a6=a5+d=(3+42)+2a_6 = a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2}

=3+52= 3 + 5\sqrt{2}

a7=a6+d=(3+52)+2a_7 = a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2}

=3+62= 3 + 6\sqrt{2}

Next three terms: 3+42,3+52,3+62\boxed{\text{Next three terms: } 3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}}

Step 6 — Check sequence (vi)

Let's find the differences between terms.

a2−a1=0.22−0.2a_2 - a_1 = 0.22 - 0.2

=0.02= 0.02

a3−a2=0.222−0.22a_3 - a_2 = 0.222 - 0.22

=0.002= 0.002

The differences are not the same.

Not an AP\boxed{\text{Not an AP}}

Step 7 — Check sequence (vii)

Let's find the differences between terms.

a2−a1=−4−0a_2 - a_1 = -4 - 0

=−4= -4

a3−a2=−8−(−4)a_3 - a_2 = -8 - (-4)

=−8+4= -8 + 4

=−4= -4

a4−a3=−12−(−8)a_4 - a_3 = -12 - (-8)

=−12+8= -12 + 8

=−4= -4

The common difference is constant.

d=−4\boxed{d = -4}

Let's find the next three terms.

a5=a4+d=−12+(−4)a_5 = a_4 + d = -12 + (-4)

=−16= -16

a6=a5+d=−16+(−4)a_6 = a_5 + d = -16 + (-4)

=−20= -20

a7=a6+d=−20+(−4)a_7 = a_6 + d = -20 + (-4)

=−24= -24

Next three terms: −16,−20,−24\boxed{\text{Next three terms: } -16, -20, -24}

Step 8 — Check sequence (viii)

Let's find the differences between terms.

a2−a1=−12−(−12)a_2 - a_1 = -\frac{1}{2} - (-\frac{1}{2})

=0= 0

a3−a2=−12−(−12)a_3 - a_2 = -\frac{1}{2} - (-\frac{1}{2})

=0= 0

a4−a3=−12−(−12)a_4 - a_3 = -\frac{1}{2} - (-\frac{1}{2})

=0= 0

The common difference is constant.

d=0\boxed{d = 0}

Let's find the next three terms.

a5=a4+d=−12+0a_5 = a_4 + d = -\frac{1}{2} + 0

=−12= -\frac{1}{2}

a6=a5+d=−12+0a_6 = a_5 + d = -\frac{1}{2} + 0

=−12= -\frac{1}{2}

a7=a6+d=−12+0a_7 = a_6 + d = -\frac{1}{2} + 0

=−12= -\frac{1}{2}

Next three terms: −12,−12,−12\boxed{\text{Next three terms: } -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}}

Step 9 — Check sequence (ix)

Let's find the differences between terms.

a2−a1=3−1a_2 - a_1 = 3 - 1

=2= 2

a3−a2=9−3a_3 - a_2 = 9 - 3

=6= 6

The differences are not the same.

Not an AP\boxed{\text{Not an AP}}

Step 10 — Check sequence (x)

Let's find the differences between terms.

a2−a1=2a−aa_2 - a_1 = 2a - a

=a= a

a3−a2=3a−2aa_3 - a_2 = 3a - 2a

=a= a

a4−a3=4a−3aa_4 - a_3 = 4a - 3a

=a= a

The common difference is constant.

d=a\boxed{d = a}

Let's find the next three terms.

a5=a4+d=4a+aa_5 = a_4 + d = 4a + a

=5a= 5a

a6=a5+d=5a+aa_6 = a_5 + d = 5a + a

=6a= 6a

a7=a6+d=6a+aa_7 = a_6 + d = 6a + a

=7a= 7a

Next three terms: 5a,6a,7a\boxed{\text{Next three terms: } 5a, 6a, 7a}

Step 11 — Check sequence (xi)

Let's find the differences between terms.

a2−a1=a2−aa_2 - a_1 = a^2 - a

=a(a−1)= a(a-1)

a3−a2=a3−a2a_3 - a_2 = a^3 - a^2

=a2(a−1)= a^2(a-1)

The differences are not generally the same. For example, if a=2a=2, a(a−1)=2(1)=2a(a-1) = 2(1) = 2 and a2(a−1)=4(1)=4a^2(a-1) = 4(1) = 4.

Not an AP\boxed{\text{Not an AP}}

Step 12 — Check sequence (xii)

Let's simplify the terms first.

8=4×2=22\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}

18=9×2=32\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}

32=16×2=42\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}

The sequence is 2,22,32,42,…\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots. Let's find the differences between terms.

a2−a1=22−2a_2 - a_1 = 2\sqrt{2} - \sqrt{2}

=2= \sqrt{2}

a3−a2=32−22a_3 - a_2 = 3\sqrt{2} - 2\sqrt{2}

=2= \sqrt{2}

a4−a3=42−32a_4 - a_3 = 4\sqrt{2} - 3\sqrt{2}

=2= \sqrt{2}

The common difference is constant.

d=2\boxed{d = \sqrt{2}}

Let's find the next three terms.

a5=a4+d=42+2a_5 = a_4 + d = 4\sqrt{2} + \sqrt{2}

=52=50= 5\sqrt{2} = \sqrt{50}

a6=a5+d=52+2a_6 = a_5 + d = 5\sqrt{2} + \sqrt{2}

=62=72= 6\sqrt{2} = \sqrt{72}

a7=a6+d=62+2a_7 = a_6 + d = 6\sqrt{2} + \sqrt{2}

=72=98= 7\sqrt{2} = \sqrt{98}

Next three terms: 50,72,98\boxed{\text{Next three terms: } \sqrt{50}, \sqrt{72}, \sqrt{98}}

Step 13 — Check sequence (xiii)

Let's simplify the terms.

9=3\sqrt{9} = 3

12=4×3=23\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}

The sequence is 3,6,3,23,…\sqrt{3}, \sqrt{6}, 3, 2\sqrt{3}, \dots. Let's find the differences between terms.

a2−a1=6−3a_2 - a_1 = \sqrt{6} - \sqrt{3}

a3−a2=3−6a_3 - a_2 = 3 - \sqrt{6}

The differences are not the same. 6−3≈0.717\sqrt{6} - \sqrt{3} \approx 0.717 and 3−6≈0.5513 - \sqrt{6} \approx 0.551.

Not an AP\boxed{\text{Not an AP}}

Step 14 — Check sequence (xiv)

Let's write out the terms.

12=11^2 = 1

32=93^2 = 9

52=255^2 = 25

72=497^2 = 49

The sequence is 1,9,25,49,…1, 9, 25, 49, \dots. Let's find the differences between terms.

a2−a1=9−1a_2 - a_1 = 9 - 1

=8= 8

a3−a2=25−9a_3 - a_2 = 25 - 9

=16= 16

The differences are not the same.

Not an AP\boxed{\text{Not an AP}}

Step 15 — Check sequence (xv)

Let's write out the terms.

12=11^2 = 1

52=255^2 = 25

72=497^2 = 49

The sequence is 1,25,49,73,…1, 25, 49, 73, \dots. Let's find the differences between terms.

a2−a1=25−1a_2 - a_1 = 25 - 1

=24= 24

a3−a2=49−25a_3 - a_2 = 49 - 25

=24= 24

a4−a3=73−49a_4 - a_3 = 73 - 49

=24= 24

The common difference is constant.

d=24\boxed{d = 24}

Let's find the next three terms.

a5=a4+d=73+24a_5 = a_4 + d = 73 + 24

=97= 97

a6=a5+d=97+24a_6 = a_5 + d = 97 + 24

=121= 121

a7=a6+d=121+24a_7 = a_6 + d = 121 + 24

=145= 145

Next three terms: 97,121,145\boxed{\text{Next three terms: } 97, 121, 145}

Answer

(i) Not an AP (ii) AP, d=12d = \frac{1}{2}, Next three terms: 4,92,54, \frac{9}{2}, 5 (iii) AP, d=−2.0d = -2.0, Next three terms: −9.2,−11.2,−13.2-9.2, -11.2, -13.2 (iv) AP, d=4d = 4, Next three terms: 6,10,146, 10, 14 (v) AP, d=2d = \sqrt{2}, Next three terms: 3+42,3+52,3+623 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2} (vi) Not an AP (vii) AP, d=−4d = -4, Next three terms: −16,−20,−24-16, -20, -24 (viii) AP, d=0d = 0, Next three terms: −12,−12,−12-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} (ix) Not an AP (x) AP, d=ad = a, Next three terms: 5a,6a,7a5a, 6a, 7a (xi) Not an AP (xii) AP, d=2d = \sqrt{2}, Next three terms: 50,72,98\sqrt{50}, \sqrt{72}, \sqrt{98} (xiii) Not an AP (xiv) Not an AP (xv) AP, d=24d = 24, Next three terms: 97,121,14597, 121, 145

More questions in Exercise 5.1

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 ₹ 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 14\frac{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.

Q2
  1. Write first four terms of the AP, when the first term aa and the common difference dd are given as follows:

(i) a=10,d=10a = 10, \quad d = 10

(ii) a=−2,d=0a = -2, \quad d = 0

(iii) a=4,d=−3a = 4, \quad d = -3

(iv) a=−1,d=12a = -1, \quad d = \frac{1}{2}

(v) a=−1.25,d=−0.25a = -1.25, \quad d = -0.25

Q3
  1. For the following APs, write the first term and the common difference:

(i) 3,1,−1,−3,…3, 1, -1, -3, \dots

(ii) −5,−1,3,7,…-5, -1, 3, 7, \dots

(iii) 13,53,93,133,…\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots

(iv) 0.6,1.7,2.8,3.9,…0.6, 1.7, 2.8, 3.9, \dots

Q4
  1. Which of the following are APs ? If they form an AP, find the common difference dd and write three more terms.

(i) 2,4,8,16,…2, 4, 8, 16, \dots

(ii) 2,52,3,72,…2, \frac{5}{2}, 3, \frac{7}{2}, \dots

(iii) −1.2,−3.2,−5.2,−7.2,…-1.2, -3.2, -5.2, -7.2, \dots

(iv) −10,−6,−2,2,…-10, -6, -2, 2, \dots

(v) 3,3+2,3+22,3+32,…3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots

(vi) 0.2,0.22,0.222,0.2222,…0.2, 0.22, 0.222, 0.2222, \dots

(vii) 0,−4,−8,−12,…0, -4, -8, -12, \dots

(viii) −12,−12,−12,−12,…-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots

(ix) 1,3,9,27,…1, 3, 9, 27, \dots

(x) a,2a,3a,4a,…a, 2a, 3a, 4a, \dots

(xi) a,a2,a3,a4,…a, a^2, a^3, a^4, \dots

(xii) 2,8,18,32,…\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots

(xiii) 3,6,9,12,…\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots

(xiv) 12,32,52,72,…1^2, 3^2, 5^2, 7^2, \dots

(xv) 12,52,72,73,…1^2, 5^2, 7^2, 73, \dots

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