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, \dfrac{5}{2}, 3, \dfrac{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,…-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{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
Understand the Question

A sequence of numbers a1,a2,a3,…a_1, a_2, a_3, \dots forms an Arithmetic Progression (AP) if the difference between any two consecutive terms is constant: ak+1−ak=dfor all k≥1a_{k+1} - a_k = d \quad \text{for all } k \ge 1 where dd is called the common difference.

  • If the differences between consecutive terms are equal, the sequence is an AP.
  • The next terms can be found by repeatedly adding the common difference dd: an+1=an+da_{n+1} = a_n + d

(i) 2,4,8,16,…2, 4, 8, 16, \dots

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=4−2=2a3−a2=8−4=4\begin{aligned} a_2 - a_1 &= 4 - 2 = 2 \\ a_3 - a_2 &= 8 - 4 = 4 \end{aligned}

Since a3−a2≠a2−a1a_3 - a_2 \neq a_2 - a_1, the difference is not constant.

Answer

(i) It is not an AP.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=52−2=12a3−a2=3−52=12a4−a3=72−3=12\begin{aligned} a_2 - a_1 &= \dfrac{5}{2} - 2 = \dfrac{1}{2} \\[0.6em] a_3 - a_2 &= 3 - \dfrac{5}{2} = \dfrac{1}{2} \\[0.6em] a_4 - a_3 &= \dfrac{7}{2} - 3 = \dfrac{1}{2} \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=12d = \dfrac{1}{2}.

Step 2 · Find the Next Three Terms

Add d=12d = \dfrac{1}{2} to find the next three terms

a5=a4+d=72+12=82=4a6=a5+d=4+12=92a7=a6+d=92+12=102=5\begin{aligned} a_5 &= a_4 + d = \dfrac{7}{2} + \dfrac{1}{2} = \dfrac{8}{2} = 4 \\[0.6em] a_6 &= a_5 + d = 4 + \dfrac{1}{2} = \dfrac{9}{2} \\[0.6em] a_7 &= a_6 + d = \dfrac{9}{2} + \dfrac{1}{2} = \dfrac{10}{2} = 5 \end{aligned}
Answer

(ii) It is an AP with d=12d = \dfrac{1}{2}. The next three terms are 4,92,54, \dfrac{9}{2}, 5.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=−3.2−(−1.2)=−3.2+1.2=−2.0a3−a2=−5.2−(−3.2)=−5.2+3.2=−2.0a4−a3=−7.2−(−5.2)=−7.2+5.2=−2.0\begin{aligned} a_2 - a_1 &= -3.2 - (-1.2) = -3.2 + 1.2 = -2.0 \\ a_3 - a_2 &= -5.2 - (-3.2) = -5.2 + 3.2 = -2.0 \\ a_4 - a_3 &= -7.2 - (-5.2) = -7.2 + 5.2 = -2.0 \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=−2.0d = -2.0.

Step 2 · Find the Next Three Terms

Add d=−2.0d = -2.0 to find the next three terms

a5=a4+d=−7.2+(−2.0)=−9.2a6=a5+d=−9.2+(−2.0)=−11.2a7=a6+d=−11.2+(−2.0)=−13.2\begin{aligned} a_5 &= a_4 + d = -7.2 + (-2.0) = -9.2 \\ a_6 &= a_5 + d = -9.2 + (-2.0) = -11.2 \\ a_7 &= a_6 + d = -11.2 + (-2.0) = -13.2 \end{aligned}
Answer

(iii) It is an AP with d=−2.0d = -2.0. The next three terms are −9.2,−11.2,−13.2-9.2, -11.2, -13.2.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=−6−(−10)=−6+10=4a3−a2=−2−(−6)=−2+6=4a4−a3=2−(−2)=2+2=4\begin{aligned} a_2 - a_1 &= -6 - (-10) = -6 + 10 = 4 \\ a_3 - a_2 &= -2 - (-6) = -2 + 6 = 4 \\ a_4 - a_3 &= 2 - (-2) = 2 + 2 = 4 \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=4d = 4.

Step 2 · Find the Next Three Terms

Add d=4d = 4 to find the next three terms

a5=a4+d=2+4=6a6=a5+d=6+4=10a7=a6+d=10+4=14\begin{aligned} a_5 &= a_4 + d = 2 + 4 = 6 \\ a_6 &= a_5 + d = 6 + 4 = 10 \\ a_7 &= a_6 + d = 10 + 4 = 14 \end{aligned}
Answer

(iv) It is an AP with d=4d = 4. The next three terms are 6,10,146, 10, 14.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=(3+2)−3=2a3−a2=(3+22)−(3+2)=2a4−a3=(3+32)−(3+22)=2\begin{aligned} a_2 - a_1 &= (3 + \sqrt{2}) - 3 = \sqrt{2} \\ a_3 - a_2 &= (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = \sqrt{2} \\ a_4 - a_3 &= (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = \sqrt{2} \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=2d = \sqrt{2}.

Step 2 · Find the Next Three Terms

Add d=2d = \sqrt{2} to find the next three terms

a5=a4+d=(3+32)+2=3+42a6=a5+d=(3+42)+2=3+52a7=a6+d=(3+52)+2=3+62\begin{aligned} a_5 &= a_4 + d = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2} \\ a_6 &= a_5 + d = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2} \\ a_7 &= a_6 + d = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2} \end{aligned}
Answer

(v) It is an AP with d=2d = \sqrt{2}. The next three terms are 3+42,3+52,3+623 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=0.22−0.2=0.02a3−a2=0.222−0.22=0.002\begin{aligned} a_2 - a_1 &= 0.22 - 0.2 = 0.02 \\ a_3 - a_2 &= 0.222 - 0.22 = 0.002 \end{aligned}

Since a3−a2≠a2−a1a_3 - a_2 \neq a_2 - a_1, the difference is not constant.

Answer

(vi) It is not an AP.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=−4−0=−4a3−a2=−8−(−4)=−8+4=−4a4−a3=−12−(−8)=−12+8=−4\begin{aligned} a_2 - a_1 &= -4 - 0 = -4 \\ a_3 - a_2 &= -8 - (-4) = -8 + 4 = -4 \\ a_4 - a_3 &= -12 - (-8) = -12 + 8 = -4 \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=−4d = -4.

Step 2 · Find the Next Three Terms

Add d=−4d = -4 to find the next three terms

a5=a4+d=−12+(−4)=−16a6=a5+d=−16+(−4)=−20a7=a6+d=−20+(−4)=−24\begin{aligned} a_5 &= a_4 + d = -12 + (-4) = -16 \\ a_6 &= a_5 + d = -16 + (-4) = -20 \\ a_7 &= a_6 + d = -20 + (-4) = -24 \end{aligned}
Answer

(vii) It is an AP with d=−4d = -4. The next three terms are −16,−20,−24-16, -20, -24.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=−12−(−12)=0a3−a2=−12−(−12)=0a4−a3=−12−(−12)=0\begin{aligned} a_2 - a_1 &= -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0 \\[0.6em] a_3 - a_2 &= -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0 \\[0.6em] a_4 - a_3 &= -\dfrac{1}{2} - \left(-\dfrac{1}{2}\right) = 0 \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=0d = 0.

Step 2 · Find the Next Three Terms

Add d=0d = 0 to find the next three terms

a5=a4+d=−12+0=−12a6=a5+d=−12+0=−12a7=a6+d=−12+0=−12\begin{aligned} a_5 &= a_4 + d = -\dfrac{1}{2} + 0 = -\dfrac{1}{2} \\[0.6em] a_6 &= a_5 + d = -\dfrac{1}{2} + 0 = -\dfrac{1}{2} \\[0.6em] a_7 &= a_6 + d = -\dfrac{1}{2} + 0 = -\dfrac{1}{2} \end{aligned}
Answer

(viii) It is an AP with d=0d = 0. The next three terms are −12,−12,−12-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}.

(ix) 1,3,9,27,…1, 3, 9, 27, \dots

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=3−1=2a3−a2=9−3=6\begin{aligned} a_2 - a_1 &= 3 - 1 = 2 \\ a_3 - a_2 &= 9 - 3 = 6 \end{aligned}

Since a3−a2≠a2−a1a_3 - a_2 \neq a_2 - a_1, the difference is not constant.

Answer

(ix) It is not an AP.

(x) a,2a,3a,4a,…a, 2a, 3a, 4a, \dots

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=2a−a=aa3−a2=3a−2a=aa4−a3=4a−3a=a\begin{aligned} a_2 - a_1 &= 2a - a = a \\ a_3 - a_2 &= 3a - 2a = a \\ a_4 - a_3 &= 4a - 3a = a \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=ad = a.

Step 2 · Find the Next Three Terms

Add d=ad = a to find the next three terms

a5=a4+d=4a+a=5aa6=a5+d=5a+a=6aa7=a6+d=6a+a=7a\begin{aligned} a_5 &= a_4 + d = 4a + a = 5a \\ a_6 &= a_5 + d = 5a + a = 6a \\ a_7 &= a_6 + d = 6a + a = 7a \end{aligned}
Answer

(x) It is an AP with d=ad = a. The next three terms are 5a,6a,7a5a, 6a, 7a.

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

Step 1 · Check Differences Between Consecutive Terms

Calculate the differences between consecutive terms

a2−a1=a2−a=a(a−1)a3−a2=a3−a2=a2(a−1)\begin{aligned} a_2 - a_1 &= a^2 - a = a(a - 1) \\ a_3 - a_2 &= a^3 - a^2 = a^2(a - 1) \end{aligned}

Since a2(a−1)≠a(a−1)a^2(a - 1) \neq a(a - 1) for general values of aa, the difference is not constant.

Answer

(xi) It is not an AP.

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

Step 1 · Simplify Terms and Check Differences

Simplify the square roots

8=4×2=2218=9×2=3232=16×2=42\begin{aligned} \sqrt{8} &= \sqrt{4 \times 2} = 2\sqrt{2} \\[0.3em] \sqrt{18} &= \sqrt{9 \times 2} = 3\sqrt{2} \\[0.3em] \sqrt{32} &= \sqrt{16 \times 2} = 4\sqrt{2} \end{aligned}

The sequence is 2,22,32,42,…\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots.

Calculate the differences between consecutive terms

a2−a1=22−2=2a3−a2=32−22=2a4−a3=42−32=2\begin{aligned} a_2 - a_1 &= 2\sqrt{2} - \sqrt{2} = \sqrt{2} \\ a_3 - a_2 &= 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \\ a_4 - a_3 &= 4\sqrt{2} - 3\sqrt{2} = \sqrt{2} \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=2d = \sqrt{2}.

Step 2 · Find the Next Three Terms

Add d=2d = \sqrt{2} to find the next three terms

a5=a4+d=42+2=52=25×2=50a6=a5+d=52+2=62=36×2=72a7=a6+d=62+2=72=49×2=98\begin{aligned} a_5 &= a_4 + d = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50} \\[0.3em] a_6 &= a_5 + d = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72} \\[0.3em] a_7 &= a_6 + d = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98} \end{aligned}
Answer

(xii) It is an AP with d=2d = \sqrt{2}. The next three terms are 50,72,98\sqrt{50}, \sqrt{72}, \sqrt{98}.

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

Step 1 · Check Differences Between Consecutive Terms

Simplify the terms 9=3,12=23\sqrt{9} = 3, \quad \sqrt{12} = 2\sqrt{3}

Calculate the differences between consecutive terms

a2−a1=6−3a3−a2=3−6\begin{aligned} a_2 - a_1 &= \sqrt{6} - \sqrt{3} \\[0.3em] a_3 - a_2 &= 3 - \sqrt{6} \end{aligned}

Since 6−3≠3−6\sqrt{6} - \sqrt{3} \neq 3 - \sqrt{6}, the difference is not constant.

Answer

(xiii) It is not an AP.

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

Step 1 · Evaluate Terms and Check Differences

Evaluate the terms 12=1,32=9,52=25,72=491^2 = 1, \quad 3^2 = 9, \quad 5^2 = 25, \quad 7^2 = 49

The sequence is 1,9,25,49,…1, 9, 25, 49, \dots.

Calculate the differences between consecutive terms

a2−a1=9−1=8a3−a2=25−9=16\begin{aligned} a_2 - a_1 &= 9 - 1 = 8 \\ a_3 - a_2 &= 25 - 9 = 16 \end{aligned}

Since a3−a2≠a2−a1a_3 - a_2 \neq a_2 - a_1, the difference is not constant.

Answer

(xiv) It is not an AP.

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

Step 1 · Evaluate Terms and Check Differences

Evaluate the terms 12=1,52=25,72=49,a4=731^2 = 1, \quad 5^2 = 25, \quad 7^2 = 49, \quad a_4 = 73

The sequence is 1,25,49,73,…1, 25, 49, 73, \dots.

Calculate the differences between consecutive terms

a2−a1=25−1=24a3−a2=49−25=24a4−a3=73−49=24\begin{aligned} a_2 - a_1 &= 25 - 1 = 24 \\ a_3 - a_2 &= 49 - 25 = 24 \\ a_4 - a_3 &= 73 - 49 = 24 \end{aligned}

Since ak+1−aka_{k+1} - a_k is constant, the sequence is an AP with common difference d=24d = 24.

Step 2 · Find the Next Three Terms

Add d=24d = 24 to find the next three terms

a5=a4+d=73+24=97a6=a5+d=97+24=121a7=a6+d=121+24=145\begin{aligned} a_5 &= a_4 + d = 73 + 24 = 97 \\ a_6 &= a_5 + d = 97 + 24 = 121 \\ a_7 &= a_6 + d = 121 + 24 = 145 \end{aligned}
Answer

(xv) It is an AP with d=24d = 24. The next three terms are 97,121,14597, 121, 145.

Common Mistakes
  • Checking only one difference: Checking only a2−a1a_2 - a_1 is insufficient. You must check a3−a2a_3 - a_2 and a4−a3a_4 - a_3 to ensure the difference is constant across all given terms.
  • Not simplifying radicals: In sequence (xii), treating 8−2\sqrt{8} - \sqrt{2} as not subtractable is a mistake. Simplifying radicals to 22−2=22\sqrt{2} - \sqrt{2} = \sqrt{2} reveals that the sequence is indeed an AP.
  • Assuming power patterns are never APs: In sequence (xv), 12,52,72,731^2, 5^2, 7^2, 73 gives 1,25,49,731, 25, 49, 73, which has a constant common difference of 2424, forming an AP even though part (xiv) does not.

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\dfrac{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 = \dfrac{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,…\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{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, \dfrac{5}{2}, 3, \dfrac{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,…-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{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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