NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions – Here are all the NCERT solutions for Class 10 Maths Chapter 5. This solution contains questions, answers, images, explanations of the complete Chapter 5 titled Arithmetic Progressions of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 5 Arithmetic Progressions. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions in one place.
NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions
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| Class | 10 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 5 |
| Chapter Name |
Arithmetic Progressions |
NCERT Solutions Class 10 Maths chapter 5 Arithmetic Progressions
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Question & Answer
Q.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 \(\frac{1}{4}\) 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.
Ans : (i) It can be observe that Taxi fare for \( 1^{st} \) km = 15 Taxi fare for first 2 km = 15 + 8 = 23 Taxi fare for first 3 km = 23 + 8 = 31 Taxi fare for first 4 km = 31 + 8 = 39 Clearly 15, 23, 31, 39.… forms an A.P. because every term is 8 more than the preceding term (ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes \(\frac{1}{4}\) of air remaining in the cylinder at a time. In other words, after every stroke ,only \(1-\frac{1}{4}=\frac{3}{4} t h\) part of air will remain. Therefore, volumes will be \(V\left(\frac{3 V}{4}\right) \cdot\left(\frac{3}{4} V\right)^{2} \cdot\left(\frac{3}{4} V\right)^{3} \dots\) Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P (iii) Cost of digging for first metre = 150 Cost of digging for first 2 metres = 150 + 50 = 200 Cost of digging for first 3 metres = 200 + 50 = 250 Cost of digging for first 4 metres = 250 + 50 = 300 Clearly, 150, 200, 250, 300 forms an A.P. because every term is 50 more than the preceding term. (iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be \(\mathbf{P}\left(1+\frac{r}{100}\right)^{n}\) after n years. Therefore, after every year, our money will be \(10000\left(1+\frac{8}{100}\right), 10000\left(1+\frac{8}{100}\right)^{2}, 10000\left(1+\frac{8}{100}\right)^{3}, 10000\left(1+\frac{8}{100}\right)^{4}\) Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
Q.2: 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
Ans : \( a=10, d=10 \) Let the series \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \ldots \) \( \begin{array}{l}{a_{1}=a=10} \\ {a_{2}=a_{1}+d=10+10=20} \\ {a_{3}=a_{2}+d=20+10=30} \\ {a_{4}=a_{3}+d=30+10=40} \\ {a_{5}=a_{4}+d=40+10=50}\end{array} \) Therefore , the series will be 10,20,30,40,50.. First four terms of this A.P. will be 10, 20, 30, and 40. (ii) \( a=-2, d=0 \) Let the series \( a_{1,} a_{2}, a_{3}, a_{4} \dots \) \( \begin{array}{l}{a_{1}=a=-2} \\ {a_{2}=a_{1}+d=-2+0=-2} \\ {a_{3}=a_{2}+d=-2+0=-2} \\ {a_{4}=a_{3}+d=-2+0=-2}\end{array} \) Therefore the series will be \( -2,-2,-2,-2 \ldots\) First four terms of this AP. Will be \( -2,-2,-2 \text { and }-2 \) (iii) \( a=4, d=-3 \) Let the series be \( a_{1}, a_{2}, a_{3 r} a_{4} \dots \) \( \begin{array}{l}{a_{1}=a=4} \\ {a_{2}=a_{1}+d=4-3=1} \\ {a_{3}=a_{2}+d=1-3=-2} \\ {a_{4}=a_{3}+d=-2-3=-5}\end{array} \) (iv) \(a=-1, d=\frac{1}{2} \) Let the series be \( a_{1}, a_{2}, a_{3}, a_{4} \dots \) \( \begin{array}{l}{a_{1}=a=-1} \\ {a_{2}=a_{1}+d=-1+\frac{1}{2}=-\frac{1}{2}} \\ {a_{3}=a_{2}+d=-\frac{1}{2}+\frac{1}{2}=0} \\ {a_{4}=a_{3}+d=0+\frac{1}{2}=\frac{1}{2}}\end{array} \) Clearly the series will be \( -1,-\frac{1}{2}, 0, \frac{1}{2} \) Four terms of this A.P. will be \( -1,-\frac{1}{2}, 0 \text { and } \frac{1}{2} \) (v) \( a=-1.25, d=-0.25\) Let the series be \( a_{1}, a_{2}, a_{3 r} a_{4} \dots \) \( \begin{array}{l}{a_{1}=a=-1.25} \\ {a_{2}=a_{1}+d=-1.25-0.25=-1.50} \\ {a_{3}=a_{2}+d=-1.50-0.25=-1.75}\end{array} \) \(a_{4}=a_{3}+d=-1.75-0.25=-2.00 \) Clearly the series will be \( 1.25,-1.50,-1.75,-2.00 \ldots \ldots . \) First four terms of this A.P will be \( -1.25,-1.50,-1.75 \text { and }-2.00 \)
Q.3: For the following APs, write the first term and the common difference:
\( \begin{array}{l}{\text { (i) } 3,1,-1,-3 \ldots} \\ {\text { (ii) }-5,-1,3,7 \ldots} \\ {\text { (iii) } 3, \frac{5}{3}, \frac{9}{3}, \frac{13}{3} \ldots} \\ {\text { (iv) } 0.6,1.7,2.8,3.9 \ldots}\end{array} \)
Ans : (i) \( 3,1,-1,-3 \ldots \) Here, first term, a = 3 Common difference, d = Second term — First term = 1-3=-2 (ii) \(-5,-1,3,7 \ldots \) Here, first term, a = —5 Common difference, d = Second term — First term \( =(-1)-(-5)=-1+5=4 \) (iii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3} \ldots \) Here, first term \( a=\frac{1}{3} \) Common difference, d = Second term — First term \( =\frac{5}{3}-\frac{1}{3}=\frac{4}{3} \) (iv) \( 0.6,1.7,2.8,3.9 \ldots \) Here, first term , a=0.6 Common difference, d = second term - First term \( \begin{array}{l}{=1.7-0.6} \\ {=1.1}\end{array} \)
Q.4: Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) \( 2,4,8,16, \dots \)
(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \)
(iii) \( -1.2,-3.2,-5.2,-7.2, \dots \)
(iv) \(-10,-6,-2,2, \dots \)
(v) \( 3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots \)
(vi) \( 0.2,0.22,0.222,0.2222 \)
(vii) \( 0,-4,-8,-12, \dots \)
(viii) \( -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \)
(ix) \( 1,3,9,27, \dots \)
(x) \( a, 2 a, 3 a, 4 a, \dots \)
(xi) \( a, a^{2}, a^{3}, a^{4}, \dots \)
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots \)
(xiv) \( \mathrm{1}^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots \)
(xv) \(1^{2}, 5^{2}, 7^{2}, 73, \dots \)
Ans : Missing
Q.5: Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP
Ans : (I) \( a=7, d=3, n=8, a_{n}=? \) We know that , For an A.P \( a_{n}=a+(n-1) d \) \( \begin{array}{l}{=7+(8-1) 3} \\ {=7+(7) 3} \\ {=7+21=28}\end{array}\) Hence \( a_{n}=28 \) (II) Given that \( a=-18, n=10, a_{n}=0, d=? \) We know that \( a=-18, n=10, a_{n}=0, d=? \) We know that \( \begin{array}{l}{a_{n}=a+(n-1) d} \\ {0=-18+(10-1) d} \\ {18=9 d} \\ {d=\frac{18}{9}=2}\end{array} \) Hence common difference , d=2 (III) Given that \( d=-3, n=18, a_{n}=-5 \) We know that \( \begin{array}{l}{a_{n}=a+(n-1) d} \\ {-5=a+(18-1)(-3)} \\ {-5=a+(17)(-3)} \\ {-5=a-51} \\ {a=51-5=46}\end{array} \) Hence a=46 (IV) \( a=-18.9, d=2.5, a_{n}=3.6, n=? \) We know that \( \begin{array}{l}{a_{n}=a+(n-1) d} \\ {3.6=-18.9+(n-1) 2.5} \\ {3.6+18.9=(n-1) 2.5} \\ {22.5=(n-1) 2.5} \\ {(n-1)=\frac{22.5}{2.5}} \\ {n-1=9} \\ {n=10}\end{array} \) Hence, n=10 (V) \( a=3.5, d=0, n=105, a_{n}=? \) We know that \( \begin{array}{l}{a_{n}=a+(n-1) d} \\ {a_{n}=3.5+(105-1) 0} \\ {a_{n}=3.5+104 \times 0} \\ {a_{n}=3.5} \\ {\text { Hence, } a_{n}=3.5}\end{array} \)
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