CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Exercise 5.1
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now1. 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 Rs 15 for the first km and Rs 8 for each additional km.
Answer
For a sequence to be AP, the difference of consecutive terms should remain constant and that is called the common difference of the \( A P \)Fare for 1 st \( \mathrm{km}= \) Rs. 15
Fare for \( 2 \mathrm{nd} \ \mathrm{km}= \) Fare of first km \(+ \) Additional fare for 1 km
\( =\text {Rs. } 15+8 =\text {Rs } 23 \)
Fare for \( 3 \mathrm{rd} \ \mathrm{km}= \) Fare of first km \(+ \) Fare of additional second km \(+ \) Fare of additional third km
\( =\text {Rs. } 23+8 =\text {Rs } 31 \)
Fare of n \( \mathrm{km}=15+8 \mathrm{x}(\mathrm{n}-1) \)
We multiplied by \( \mathrm{n}-1 \) because first km was fixed and for rest we are adding additional fare.
In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.
Hence, it is an AP
(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.
Answer
For a sequence to be AP, the difference of consecutive terms should remain constant and that is called the common difference of the \( A P \)Let us take initial quantity of air \( =1 \)
Hence, quantity of air removed in first step \( =\frac{1 }{ 4} \)
Remaining quantity after 1 st step
\( =1-\frac{1}{4}=\frac{3}{4} \)
Quantity removed after 2nd step \(=\) Quantity removed in first step \(\times\) initial quantity
\( =\frac{3}{4} \times \frac{1}{4}=\frac{3}{16} \)
Remaining quantity after 2nd step would be:
\( =\frac{3}{4}-\frac{3}{16}=\frac{12-3}{16}=\frac{9}{16} \)
After second step the difference between second and first and first and initial step is not the same, hence
Here a fixed number is not added to each subsequent term.
Hence, it is not an AP
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
Answer
For a sequence to be AP, the difference of consecutive terms should remain constant and that is called the common difference of the \( A P \)Cost of digging of 1 st meter \( = \) Rs 150
Cost of digging of second meter \( = \) cost of digging of first meter \(+\) cost of digging additional meter
Cost of digging of 2 nd meter \( =150+50 \)
\( =\text{Rs } 200 \)
Cost of digging of third meter \( = \) Cost of digging of first meter \(+\) cost of digging of second meter \(+\) cost of digging of third meter
Cost of digging of 3 rd meter \( =200+50 \)
\( =\text {Rs } 250 \)
Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.
Hence, it is an AP.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at \( 8 \% \) per annum.
Answer
For a sequence to be AP, the difference of consecutive terms should remain constant and that is called the common difference of the \( A P \)Amount in the beginning \( = \) Rs. 10000
Interest at the end of 1 st year at rate of \( 8 \% \)
\(=10000 \times 8 \%\)
\(=800\)
Hence, amount at the end of 1 styear
\(=10000+800\)
\(=10800\)
Now the interest will be made at the principal taken as amount of first year, Hence
Interest at the end of 2 nd year at rate of \( 8 \% \)
\(=10800 \times 8 \%\)
\(=864\)
Thus, amount at the end of 2nd year
\( =10800+864=11664 \)
Since, each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP.
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now2. Write first four terms of the AP, when the first term a and the common difference d aregiven as follows:
(i) \( \mathrm{a}=10, \mathrm{~d}=10 \)
Answer
(i) Here, first term a1 = 10 and common difference, \( \mathrm{d}=10 \)Hence,
\(\text {2nd term } \mathrm{a}_{2}=\mathrm{a}_{1}+\mathrm{d}\)
\(=10+10\)
\(=20\)
3rd term \(a_{3} =\mathrm{a}_{1}+2 \mathrm{d} \)
\(=10+2 \times 10\)
\(=30\)
4th term \( a_{4}=a_{1}+3 d \)
\(=10+30\)
\(=40\)
Therefore,
first four terms of the AP are:
\( 10,20,30,40, \ldots \ldots \)
(ii) \( a=-2, d=0 \)
Answer
Here,First term \( \mathrm{a}=-2 \) and Common difference \( =0 \)
Therefore, first four terms of the given AP are:
\( a_{1}=-2, a_{2}=-2, a_{3}=-2 \) and \( a_{4}=-2 \)
(iii) \( a=4, d=-3 \)
Answer
Here, first term \( a_{1} =4 \) and common difference \( d=-3 \)We know that an \( =\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \), where \( \mathrm{n}= \) number of terms Thus, second term \( \mathrm{a}_{2}=\mathrm{a}+(2-1) \mathrm{d} \)
\(a_{2}=4+(2-1) \times(-3)\)
\(=4-3=1\)
\(3 \text {rd term } a_{3}=\mathrm{a}+(3-1) \mathrm{d}\)
\(=4+(3-1) \times(-3)\)
\(=4-6\)
\(=-2\)
\(4 \text {th term } \mathrm{a}_{4}=\mathrm{a}+(4-1) \mathrm{d}\)
\(=4+(4-1) \times(-3)\)
\(=4-9=-5\)
Therefore,
First four terms of given AP are:
\( 4,1,-2,-5 \)
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
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(iv) \( a=-1, d=\frac{1 }{ 2} \)
Answer
We have,1st term \( =-1 \) and \( d=\frac{1 }{ 2} \)
Hence,
\( \text {2nd term } \mathrm{a}_{2}=\mathrm{a}_{1}+\mathrm{d}\)
\(=-1+\frac{1 }{ 2}\)
\(=-\frac{1 }{ 2}\)
\(3\text {rd term } \mathrm{a}_{3}=\mathrm{a}_{1}+2 \mathrm{d}\)
\(=-1+2^{*} \frac{1 }{ 2}\)
\(=0\)
4th term \( \mathrm{a}_{4}=\mathrm{a}_{1}+3 \mathrm{d} \)
\(=-1+3^{*} \frac{1 }{ 2}\)
\(=\frac{1 }{ 2}\)
Therefore,
The four terms of A.P. are \( -1,-\frac{1 }{ 2},0,\frac{1 }{ 2} \)
(v) \( \mathrm{a}=-1.25, \mathrm{~d}=-0.25 \)
Answer
We have\( 1 \text {st term }=-1.25 \text { and } d=-0.25\)
\(2 \text {nd term } a_{2}=\mathrm{a}+\mathrm{d}\)
\(=-1.25-0.25\)
\(=-1.5\)
\(3 \text{rd term } a_{3}=\mathrm{a}+2 \mathrm{d}\)
\(=-1.25+2 \times(-0.25)\)
\(=-1.25-0.5\)
\(=-1.75\)
4th term \( \mathrm{a}_{4}=\mathrm{a}+3 \mathrm{d} \)
\(=-1.25+3 \times(-0.25)\)
\(=-2\)
Therefore, first four terms of the A.P. are: \( -1.25,-1.5,-1.75 \) and \(-2 \).
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now3. For the following APs, write the first term and the common difference:
(i) \( 3,1,-1,-3, \ldots \)
Answer
Here, first term \( \mathrm{a}=3 \)Now,
The Common difference of the A.P. can be calculated as:
\(a_{4}-a_{3}\)
\(=-3-(-1)\)
\(=-3+1\)
\(=-2\)
\(a_{3}-a_{2}\)
\(=-1-1\)
\(=-2\)
\(a_{2}-a_{1}\)
\(=1-3\)
\(=-2\)
Now, here \( a_{k+1}-a_{k}=-2 \) for all values of \( k \)
Hence, first term \( =3 \) and common difference \( =-2 \)
(ii) \( -5,-1,3,7 \)
Answer
\(a_{4}-a_{3}\)\(=7-3\)
\(=4\)
\(a_{3}-a_{2}\)
\(=3-(-1)\)
\(=3+1\)
\(=4\)
\(a_{2}-a_{1}\)
\(=-1-(-5)\)
\(=-1+5\)
\(=4\)
Now, here, \( a_{k+1}-a_{k}=4 \) for all values of \( k \)
Therefore, first term \( =-5 \) and common difference \( =4 \)
(iii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{5}, \frac{13}{3}, \ldots \).
Answer
From the question,\(a_{4}-a_{3}\)
\(=\frac{13}{3}-\frac{9}{3}\)
\(=\frac{4}{3}\)
\(a_{3}-a_{2}\)
\(=\frac{9}{3}-\frac{5}{3}\)
\(=\frac{4}{3}\)
\(a_{2}-a_{1}\)
\( =\frac{4}{3} \)
Now, \( a_{k+1}-a_{k}=\frac{4}{3} \) for all values of \( k \)
Therefore,
First term \( =\frac{1 }{ 3 }\) and common difference \( =\frac{4 }{ 3} \)
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now(iv) \( 0.6,1.7,2.8,3.9, \ldots \)
Answer
\(\mathrm{a}_{4}-\mathrm{a}_{3}\)\(=3.9-2.8\)
\(=1.1\)
\(\mathrm{a}_{3}-\mathrm{a}_{2}\)
\(=2.8-1.7\)
\(=1.1\)
\(\mathrm{a}_{2}-\mathrm{a}_{1}\)
\(=1.7-0.6\)
\(=1.1\)
Now, here, \( a_{k+1}-a_{k}=1.1 \) for all values of \( k \)
Therefore,
First term \( =0.6 \) and common difference \( =1.1 \)
4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) \( 2,4,8,16 \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APif \( a_{k+1}-a_{k} \) is same for different values of \( k \) then the series is an AP.
We have, \( a_{1}=2, a_{2}=4, a_{3}=8 \) and \( a_{4}=16 \)
\(a_{4}-a_{3}=16-8=8\)
\(a_{3}-a_{2}=8-4=4\)
\(a_{2}-a_{1}=4-2=2\)
Here, \( a_{k+1}-a_{k} \) is not same for all values of \( k \).
Hence, the given series is not an AP.
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
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(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2} \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APAs per the question:
\(\mathrm{a}_{1}=2\)
\(\mathrm{a}_{2}=\frac{5 }{ 2}\)
\(\mathrm{a}_{3}=3\)
And
\(a_{4}=\frac{7}{2}\)
\(a_{4}-a_{3}=\frac{7}{2}-3=\frac{1 }{ 2}\)
\(a_{3}-a_{2}=3-\frac{5}{2}=\frac{1 }{ 2}\)
\(a_{2}-a_{1}=\frac{5}{2}-2=\frac{1 }{ 2}\)
Now, we can observe that \( a_{k+1}-a_{k} \) is same for all values of \( k \).
Hence, it is an AP.
And, the common difference \( =\frac{1 }{ 2} \)
Next three terms of this series are:
\(a_{5}=a+4 d\)
\(=2+4^{*} \frac{1 }{ 2}\)
\(=4\)
\(a_{6}=a+5 d\)
\(=2+5^{*} \frac{1 }{ 2}\)
\(=\frac{9}{2}\)
\(a_{7}=\mathrm{a}+6 d\)
\(=2+6^{*} \frac{1 }{ 2}\)
\(=5\)
Hence,
The next three terms of the AP are: \( 4,\frac{9 }{ 2} \) and \(5\)
(iii) \( -1.2,-3.2,=5.2,-78.2, \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( \mathrm{a}_{4}-\mathrm{a}_{3}=-7.2+5.2=-2\)
\(\mathrm{a}_{3}-\mathrm{a}_{2}=-5.2+3.2=-2\)
\(\mathrm{a}_{2}-\mathrm{a}_{1}=-3.2+1.2=-2\)
In this, \( a_{k+1}-a_{k} \) is same for all values of \( k \).
Hence, the given series is an AP.
Common difference \( =-2 \)
Next three terms of the series are:
\(\mathrm{a}_{5}=\mathrm{a}+4 \mathrm{d}\)
\(=-1.2+4 \times(-2)\)
\(=-1.2-8\)
\(=-9.2\)
\(a_{6}=a+5 d\)
\(=-1.2+5 \times(-2)\)
\(=-1.2-10\)
\(=-11.2\)
\(a_{7}=a+6 d\)
\(=-1.2+6 \times(-2)\)
\(=-1.2-12\)
\(=-13.2\)
Next three terms of AP are: \( -9.2,-11.2 \) and \(-13.2\)
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now(iv) \( -10,-6,-2,2, \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( a_{4}-a_{3}=2+2=4 \)
\(a_{3}-a_{2}=-2+6=4\)
\(a_{2}-a_{1}=-6+10=4\)
Here, \( a_{k+1}-a_{k} \) is same for all values of \( k \)
Hence, the given series is an AP.
Common difference \( =4 \)
Next three terms of the AP are:
\( \mathrm{a}_{5}=\mathrm{a}+4 \mathrm{d}\)
\(=-10+4 \times 4\)
\(=-10+16=6\)
\(a_{6}=a+5 d\)
\(=-10+5 \times 4\)
\(=-10+20=10\)
\(a_{7}=a+6 d\)
\(=-10+6 \times 4\)
\(=-10+24=14\)
Next three terms of AP are: 6,10 and 14.
(v) \( 3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2} \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( \mathrm{a}_{4}-\mathrm{a}_{3}=3+3 \sqrt{ 2} -3-2 \sqrt{ 2} =\sqrt{ 2 } \) \(a_{3}-a_{2}=3+2 \sqrt{2 } -3-\sqrt{ 2} =\sqrt{2 } \)
\(a_{2}-a_{1}=3+\sqrt{ 2 }-3=\sqrt{2 } \)
Here, \( a_{k+1}-a_{k} \) is same for all values of \( k \)
Hence, the given series is an AP.
Common difference \( =\sqrt{ 2} \)
Next three terms of the AP are
\(a_{6}=a+5 d=3+5 \sqrt{ 2 }\)
\(a_{7}=a+6 d=3+6 \sqrt{ 2} \)
Next three terms of AP are: \( 3+4 \sqrt{2 } ,3+5 \sqrt{2 } \) and \( 3+6 \sqrt{2 } \)
(vi) \( 0.2,0.22,0.222,0.2222, \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\(a_{4}-a_{3}=0.2222-0.222=0.0002\)
\(a_{3}-a_{2}=0.222-0.22=0.002\)
\(a_{2}-a_{1}=0.22-0.2=0.02\)
Here, \( a_{k+1}-a_{k} \) is not same for all values of \( k \)
Hence, the given series is not an AP.
(vii) \( 0,-4,-8,-12, ...\)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APHere; \( a_{4}-a_{3}=-12+8=-4 \)
\( a_{3}-a_{2}=-8+4=-4 \)
\( a_{2}-a_{1}=-4-0=-4 \)
Since \( a_{k+1}-a_{k} \) is same for all values of \( k \).
Hence, this is an AP.
The next three terms can be calculated as follows:
\( a_{5}=a+4 d=0+4(-4)=-16 \)
\( a_{6}=a+5 d=0+5(-4)=-20 \)
\( a_{7}=a+6 d=0+6(-4)=-24 \)
Thus, next three terms are; \( -16,-20 \) and \(-24\)
(viii) \( -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APHere, it is clear that \( d=0 \)
Since \( a_{k+1}-a_{k} \) is same for all values of \( k \).
Hence, it is an AP.
The next three terms will be same, i.e. \( -\frac{1 }{ 2} \)
(ix) \( 1,3,9,27 \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( \mathrm{a}_{4}-\mathrm{a}_{3}=27-9=18 \)
\( a_{3}-a_{2}=9-3=6 \)
\( \mathrm{a}_{2}-\mathrm{a}_{1}=3-1=2 \)
Since \( a_{k+1}-a_{k} \) is not same for all values of \( k \).
Hence, it is not an AP.
CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Download the Math Ninja App Now(x) \(a, 2a, 3a, 4 a, \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( a_{4}-a_{3}=4 a-3 a=a \)
\( \mathrm{a}_{3}-\mathrm{a}_{2}=3 \mathrm{a}-2 \mathrm{a}=\mathrm{a} \)
\( \mathrm{a}_{2}-\mathrm{a}_{1}=2 \mathrm{a}-\mathrm{a}=\mathrm{a} \)
Since \( a_{k+1}-a_{k} \) is same for all values of \( k \).
Hence, it is an AP.
Next three terms are:
\( a_{5}=a+4 d=a+4 a=5 a \)
\(a_{6}=a+5 d=a+5 a=6 a\)
\(a_{7}=a+6 d=a+6 a=7 a\)
Next three terms are; 5a, 6a and 7a.
(xi) \( a, a^{2}, a^{3}, a^{4}, \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APHere, the exponent is increasing in each subsequent term.
\(a_{4}=a^{4}, a_{3}=a^{3}, a_{2}=a^{2}, a_{1}=a\)
\(a_{4}-a_{3}=a^{4}-a^{3}\)
\(a_{3}-a_{2}=a^{3}-a^{2}\)
since, the difference is not same,
Since \( a_{k+1}-a_{k} \) is not same for all values of \( k \).
Hence, it is not an AP
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}... \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APDifferent terms of this AP can also be written as follows: \(\sqrt{2}, 2 \sqrt{ 2} ,3 \sqrt{2 } ,4 \sqrt{2}, \ldots \ldots \ldots\)
\(a_{4}-a_{3}=4 \sqrt{ 2} -3 \sqrt{2}=\sqrt{2}\)
\(a_{3}-a_{2}=3 \sqrt{ 2} -2 \sqrt{2 } =\sqrt{2}\)
\(a_{2}-a_{1}=2 \sqrt{2 } -\sqrt{ 2} =\sqrt{2 } \)
Since \( a k+1-a k \) is same for all values of \( k \).
Hence, it is an AP.
Next three terms can be calculated as follows:
\(a_{5}=a+4 d=\sqrt{2 } +4 \sqrt{ 2} =5 \sqrt{2 } \)
\(a_{6}=a+5 d=\sqrt{2 } +5 \sqrt{2 } =6 \sqrt{ 2} \)
\(a_{7}=a+6 d=\sqrt{2 } +6 \sqrt{2 } =7 \sqrt{2 } \)
Next three terms are; \( 5 \sqrt{2 } ,6 \sqrt{2 } \) and \( 7 \sqrt{2 } \)
(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12} \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( a_{4}-a_{3}=\sqrt{12 } -\sqrt{ 9} =2 \sqrt{3 } -3\)
\((\sqrt{ 12} =\sqrt{ 2} \times 2 \times 3=2 \sqrt{ 3} )\)
\(a_{3}-a_{2}=\sqrt{9 } -\sqrt{ 6} =3-\sqrt{6 } \)
\(a_{2}-a_{1}=\sqrt{6 } -\sqrt{ 3 }\)
Since \( a_{k+1}-a_{k} \) is not same for all values of \( k \).
Hence, it is not an AP
(xiv) \( 1^{2}, 3^{2}, 5^{2}, 7^{2} \ldots \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of APThe given terms can be written as follows:
\( 1,9,25,49, \ldots \)
Here, \( a_{4}-a_{3}=49-25=24 \)
\( \mathrm{a}_{3}-\mathrm{a}_{2}=25-9=16 \)
\( \mathrm{a}_{2}-\mathrm{a}_{1}=9-1=8 \)
Since \( a_{k+1}-a_{k} \) is not same for all values of \( k \).
Hence, it is not an AP.
(xv) \( 1^{2}, 5^{2}, 7^{2}, 73 \)
Answer
For a sequence to be an AP, the difference between two consecutive terms remains the same, that is called the common difference of AP\( 1^{2}, 5^{2}, 7^{2}, 73 \ldots \).
\( a=1 d=5^{2}-1=25-1=24 \)
\( \mathrm{d}=7^{2}-5^{2}=49-25=24 \)
\( \mathrm{d}=74-49=24 \) As, common difference is same. The sequence is in A.P.