NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 - Arithmetic Progressions

Q1 (i) In which of the following situations does the list of numbers involved make an arithmetic progression, and why? The taxi fare after each km when the fare is $\small Rs\hspace{1mm}15$ for the first km and $\small Rs\hspace{1mm}8$ for each additional $\small km$ .

Answer:

It is given that

Fare for $1^{st} \ km$ = Rs. 15

And after that $\small Rs\hspace{1mm}8$ for each additional $\small km$

Now,

Fare for $2^{nd} \ km$ = Fare of first km + Additional fare for 1 km

= Rs. 15 + 8 = Rs 23

Fare for $3^{rd} \ km$ = Fare of first km + Fare of additional second km + Fare of additional third km

= Rs. 23 + 8= Rs 31

Fare of n km = $15 + 8 \times (n - 1)$

( We multiplied by n - 1 because the first km was fixed and for the rest, we are adding additional fare.

In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.)

Now, we can clearly see that this is an A.P. with the first term (a) = 15 and common difference (d) = 8

Q1 (ii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? The amount of air present in a cylinder when a vacuum pump removes $\small \frac{1}{4}$ of the air remaining in the cylinder at a time.

Answer:

It is given that

Vacum pump removes $\small \frac{1}{4}$ of the air remaining in the cylinder at a time

Let us take an initial quantity of air = 1

Now, the quantity of air removed in the first step = 1/4

Remaining quantity after 1^st step

$= 1-\frac{1}{4}= \frac{3}{4}$

Similarly, Quantity removed after 2^nd step = Quantity removed in first step $\times$ Remaining quantity after 1^st step

$=\frac{3}{4}\times \frac{1}{4}= \frac{3}{16}$
Now,

Remaining quantity after 2^nd step would be = Remaining quantity after 1st step - Quantity removed after 2^nd step

$=\frac{3}{4}- \frac{3}{16}= \frac{12-3}{16}= \frac{9}{16}$
Now, we can clearly see that

After the second step, the difference between the second and first and first and initial step is not the same, hence

The common difference (d) is not the same after every step

Therefore, it is not an AP.

Q1 (iii) In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? The cost of digging a well after every meter of digging, when it costs $\small Rs\hspace{1mm}150$ for the first metre and rises by $\small Rs\hspace{1mm}50$ for each subsequent meter.

Answer:

It is given that

Cost of digging of 1st meter = Rs 150

And

Rises by $\small Rs\hspace{1mm}50$ for each subsequent meter

Therefore,

Cost of digging of first 2 meters = cost of digging of first meter + cost of digging the additional meter

Cost of digging of first 2 meters = 150 + 50

= Rs 200

Similarly,

Cost of digging of first 3 meters = cost of digging of first 2 meters + cost of digging of additional meter

Cost of digging of first 3 meters = 200 + 50

= Rs 250

We can clearly see that 150, 200,250, ... is in AP with each subsequent term obtained by adding a fixed number (50) to the previous term.

Therefore, it is an AP with first term (a) = 150 and common difference (d) = 50

Q1 (iv) In which of the following situations does the list of numbers involved make an arithmetic progression, and why? The amount of money in the account every year, when $\small Rs\hspace{1mm}10000$ is deposited at compound interest at $\small 8\hspace{1mm}\%$ per annum

Answer:

Amount in the beginning = Rs. 10000

Interest at the end of 1st year at the rate of $\small 8\hspace{1mm}\%$

is $\small 8\hspace{1mm}\%$ of 10000 = $\frac{8\times 10000}{100}= 800$

Therefore, the amount at the end of 1st year will be

= 10000 + 800

= 10800

Now,

Interest at the end of 2nd year at rate of $\small 8\hspace{1mm}\%$

is $\small 8\hspace{1mm}\%$ of 10800 = $\frac{8\times 10800}{100}= 864$

Therefore, the amount at the end of 2^nd 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

Q2 (i) Write first four terms of the AP, when the first term a and the common difference d are given as follows $\small a=10,d=10$

Answer:

It is given that

$\small a=10,d=10$

Now,

$a_1= a =10$

$a_2= a_1 + d= 10 + 10 = 20$

$a_3= a_2 + d= 20 + 10 = 30$

$a_4= a_3 + d= 30 + 10 = 40$

Therefore, the first four terms of the given series are 10, 20, 30, 40.

Q2 (ii) Write first four terms of the AP when the first term a and the common difference d are given as follows: $\small a=-2,d=0$
Answer:

It is given that

$\small a=-2,d=0$

Now,

$a_1= a = -2$

$a_2= a_1 + d= -2 + 0 = -2$

$a_3= a_2 + d= -2 + 0 = -2$

$a_4= a_3 + d= -2 + 0 = -2$

Therefore, the first four terms of the given series are -2, -2, -2, -2.

Q2 (iii) Write first four terms of the AP when the first term a and the common difference d are given as follows $\small a=4,d=-3$

Answer:

It is given that

$\small a=4,d=-3$

Now,

$a_1= a =4$

$a_2= a_1 + d= 4 - 3 = 1$

$a_3= a_2 + d= 1 - 3 = -2$

$a_4= a_3 + d= -2- 3 = -5$

Therefore, the first four terms of the given series are 4, 1, -2, -5

Q2 (iv) Write the first four terms of the AP when the first term a and the common difference d are given as follows $\small a=-1,d=\frac{1}{2}$

Answer:

It is given that

$\small a=-1,d=\frac{1}{2}$

Now,

$a_1= a =-1$

$a_2= a_1 + d= -1 + \frac{1}{2} = -\frac{1}{2}$

$a_3= a_2 + d= -\frac{1}{2} + \frac{1}{2} = 0$

$a_4= a_3 + d= 0+\frac{1}{2}= \frac{1}{2}$

Therefore, the first four terms of the given series are $-1,-\frac{1}{2},0, \frac{1}{2}$

Q2 (v) Write first four terms of the AP when the first term a and the common difference d are given as follows $\small a=-1.25,d=-0.25$

Answer:

It is given that

$\small a=-1.25,d=-0.25$

Now,

$a_1= a =-1.25$

$a_2= a_1 + d= -1.25 -0.25= -1.50$

$a_3= a_2 + d= -1.50-0.25=-1.75$

$a_4= a_3 + d= -1.75-0.25=-2$

Therefore, the first four terms of the given series are -1.25, -1.50, -1.75, -2

Q3 (i) For the following APs, write the first term and the common difference $\small 3,1,-1,-3,...$

Answer:

Given AP series is

$\small 3,1,-1,-3,...$

Now, the first term of this AP series is 3

Therefore,

First term of AP series (a) = 3

Now,

$a_1=3 \ \ and \ \ a_2 = 1$

And common difference (d) = $a_2-a_1 = 1-3 = -2$

Therefore, the first term and the common difference are 3 and -2, respectively

Q3 (ii) For the following APs, write the first term and the common difference: $\small -5,-1,3,7,...$

Answer:

Given AP series is

$\small -5,-1,3,7,...$

Now, the first term of this AP series is -5

Therefore,

First-term of AP series (a) = -5

Now,

$a_1=-5 \ \ and \ \ a_2 = -1$

And common difference (d) = $a_2-a_1 = -1-(-5) = 4$

Therefore, the first term and the common difference are -5 and 4, respectively

Q3 (iii) For the following APs, write the first term and the common difference: $\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...$

Answer:

Given AP series is

$\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...$

Now, the first term of this AP series is $\frac{1}{3}$

Therefore,

The first term of AP series (a) = $\frac{1}{3}$

Now,

$a_1=\frac{1}{3} \ \ and \ \ a_2 = \frac{5}{3}$

And common difference (d) = $a_2-a_1 = \frac{5}{3}-\frac{1}{3} = \frac{5-1}{3} =\frac{4}{3}$

Therefore, the first term and the common difference is $\frac{1}{3}$ and $\frac{4}{3}$ respectively

Q3 (iv) For the following APs, write the first term and the common difference: $\small 0.6,1.7,2.8,3.9,...$

Answer:

Given the AP series is

$\small 0.6,1.7,2.8,3.9,...$

Now, the first term of this AP series is 0.6

Therefore,

First term of AP series (a) = 0.6

Now,

$a_1=0.6 \ \ and \ \ a_2 = 1.7$

And common difference (d) = $a_2-a_1 = 1.7-0.6 = 1.1$

Therefore, the first term and the common difference are 0.6 and 1.1, respectively.

Q4 (i) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small 2,4,8,12,...$

Answer:

Given series is

$\small 2,4,8,12,...$

Now,

The first term of this series is = 2

Now,

$a_1 = 2 \ \ and \ \ a_2 = 4 \ \ and \ \ a_3 = 8$

$a_2-a_1 = 4-2 = 2$

$a_3-a_2 = 8-4 = 4$

We can see that the difference between the terms is not equal

Hence, the given series is not an AP.

Q4 (ii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small 2,\frac{5}{2},3,\frac{7}{2},...$

Answer:

Given series is

$\small 2,\frac{5}{2},3,\frac{7}{2},...$

Now,

The first term of this series is = 2

Now,

$a_1 = 2 \ \ and \ \ a_2 = \frac{5}{2} \ \ and \ \ a_3 = 3 \ \ and \ \ a_4 = \frac{7}{2}$

$a_2-a_1 = \frac{5}{2}-2 = \frac{5-4}{2}=\frac{1}{2}$

$a_3-a_2 = 3-\frac{5}{2} = \frac{6-5}{2} = \frac{1}{2}$

$a_4-a_3=\frac{7}{2}-3=\frac{7-6}{2} =\frac{1}{2}$

We can clearly see that the difference between terms are equal and equal to $\frac{1}{2}$

Hence, the given series is in AP.

Now, the next three terms are

$a_5=a_4+d = \frac{7}{2}+\frac{1}{2} = \frac{8}{2}=4$

$a_6=a_5+d = 4+\frac{1}{2} = \frac{8+1}{2}=\frac{9}{2}$

$a_7=a_6+d =\frac{9}{2} +\frac{1}{2} = \frac{10}{2}=5$

Therefore, the next three terms of the given series are $4,\frac{9}{2} ,5$

Q4 (iii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small -1.2,-3.2,-5.2,-7.2,...$

Answer:

Given series is

$\small -1.2,-3.2,-5.2,-7.2,...$

Now,

The first term of this series is = -1.2

Now,

$a_1 = -1.2 \ \ and \ \ a_2 = -3.2 \ \ and \ \ a_3 = -5.2 \ \ and \ \ a_4 = -7.2$

$a_2-a_1 = -3.2-(-1.2) =-3.2+1.2=-2$

$a_3-a_2 = -5.2-(-3.2) =-5.2+3.2 = -2$

$a_4-a_3=-7.2-(-5.2)=-7.2+5.2=-2$

We can clearly see that the difference between terms are equal and equal to -2

Hence, the given series is in AP.

Now, the next three terms are

$a_5=a_4+d = -7.2-2 =-9.2$

$a_6=a_5+d = -9.2-2 =-11.2$

$a_7=a_6+d = -11.2-2 =-13.2$

Therefore, the next three terms of the given series are -9.2, -11.2, -13.2

Q4 (iv) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small -10,-6,-2,2,...$

Answer:

Given series is

$\small -10,-6,-2,2,...$

Now,

The first term of this series is = -10

Now,

$a_1 = -10 \ \ and \ \ a_2 = -6 \ \ and \ \ a_3 = -2 \ \ and \ \ a_4 = 2$

$a_2-a_1 = -6-(-10) =-6+10=4$

$a_3-a_2 = -2-(-6) =-2+6 = 4$

$a_4-a_3=2-(-2)=2+2=4$

We can clearly see that the difference between the terms are equal and equal to 4

Hence, the given series is in AP.

Now, the next three terms are

$a_5=a_4+d = 2+4 =6$

$a_6=a_5+d = 6+4=10$

$a_7=a_6+d = 10+4=14$

Therefore, the next three terms of the given series are 6,10,14

Q4 (v) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$

Answer:

Given series is

$\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$

Now,

The first term of this series is = 3

Now,

$a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2$

$a_2-a_1 = 3+\sqrt2-3= \sqrt2$

$a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2$

$a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2$

We can clearly see that the difference between terms are equal and equal to $\sqrt2$

Hence, given series is in AP

Now, the next three terms are

$a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2$

$a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2$

$a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2$

Therefore, the next three terms of given series are $3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2$

Q4 (vi) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small 0.2,0.22,0.222,0.2222,...$

Answer:

Given series is

$\small 0.2, 0.22, 0.222, 0.2222,...$

Now,

The first term to this series is = 0.2

Now,

$a_1 = 0.2 \ \ and \ \ a_2 = 0.22 \ \ and \ \ a_3 = 0.222 \ \ and \ \ a_4 = 0.2222$

$a_2-a_1 = 0.22-0.2=0.02$

$a_3-a_2 = 0.222-0.22=0.002$

We can clearly see that the difference between the terms is not equal.

Hence, the given series is not an AP

Q4 (vii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small 0,-4,-8,-12,...$

Answer:

Given series is

$\small 0,-4,-8,-12,...$

Now,

the first term to this series is = 0

Now,

$a_1 = 0 \ \ and \ \ a_2 = -4 \ \ and \ \ a_3 = -8 \ \ and \ \ a_4 = -12$

$a_2-a_1 = -4-0 =-4$

$a_3-a_2 = -8-(-4) =-8+4 = -4$

$a_4-a_3=-12-(-8)=-12+8=-4$

We can clearly see that the difference between terms are equal and equal to -4

Hence, given series is in AP

Now, the next three terms are

$a_5=a_4+d = -12-4 =-16$

$a_6=a_5+d = -16-4=-20$

$a_7=a_6+d = -20-4=-24$

Therefore, the next three terms of the given series are -16,-20,-24

Q4 (viii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$

Answer:

Given series is

$\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$

Now,

The first term to this series is = $-\frac{1}{2}$

Now,

$a_1 = -\frac{1}{2} \ \ and \ \ a_2 = -\frac{1}{2} \ \ and \ \ a_3 = -\frac{1}{2} \ \ and \ \ a_4 = -\frac{1}{2}$

$a_2-a_1 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$

$a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$

$a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$

We can clearly see that the difference between terms are equal and equal to 0

Hence, given series is in AP

Now, the next three terms are

$a_5=a_4+d = -\frac{1}{2}+0=-\frac{1}{2}$

$a_6=a_5+d = -\frac{1}{2}+0=-\frac{1}{2}$

$a_7=a_6+d = -\frac{1}{2}+0=-\frac{1}{2}$

Therefore, the next three terms of given series are $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$

Q4 (ix) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small 1,3,9,27,...$

Answer:

Given series is

$\small 1, 3, 9, 27,...$

Now,

The first term to this series is = 1

Now,

$a_1 = 1 \ \ and \ \ a_2 = 3 \ \ and \ \ a_3 = 9 \ \ and \ \ a_4 = 27$

$a_2-a_1 = 3-1=2$

$a_3-a_2 =9-3=6$

We can clearly see that the difference between terms are not equal.

Hence, given series is not an AP.

Q4 (x) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small a,2a,3a,4a,...$

Answer:

Given series is

$\small a,2a,3a,4a,...$
Now,
the first term to this series is = a
Now,
$a_1 = a \ \ and \ \ a_2 = 2a \ \ and \ \ a_3 = 3a \ \ and \ \ a_4 = 4a$
$a_2-a_1 = 2a-a =a$
$a_3-a_2 = 3a-2a =a$
$a_4-a_3=4a-3a=a$
We can clearly see that the difference between terms are equal and equal to a
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d =4a+a=5a$
$a_6=a_5+d =5a+a=6a$
$a_7=a_6+d =6a+a=7a$
Therefore, next three terms of given series are 5a,6a,7a

Q4 (xi) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small a,a^2,a^3,a^4,...$
Answer:

Given series is
$\small a,a^2,a^3,a^4,...$
Now,
the first term to this series is = a
Now,
$a_1 = a \ \ and \ \ a_2 = a^2 \ \ and \ \ a_3 = a^3 \ \ and \ \ a_4 = a^4$
$a_2-a_1 = a^2-a =a(a-1)$
$a_3-a_2 = a^3-a^2 =a^2(a-1)$

We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xii) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...$

Answer:

Given series is
$\small \sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},...$
We can rewrite it as
$\sqrt2,2\sqrt2,3\sqrt2,4\sqrt2,....$
Now,
first term to this series is = a
Now,
$a_1 = \sqrt2 \ \ and \ \ a_2 = 2\sqrt2 \ \ and \ \ a_3 = 3\sqrt2 \ \ and \ \ a_4 = 4\sqrt2$
$a_2-a_1 = 2\sqrt2-\sqrt2 =\sqrt2$
$a_3-a_2 = 3\sqrt2-2\sqrt2 =\sqrt2$
$a_4-a_3=4\sqrt2-3\sqrt2=\sqrt2$
We can clearly see that difference between terms are equal and equal to $\sqrt2$
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d =4\sqrt2+\sqrt2=5\sqrt2$
$a_6=a_5+d =5\sqrt2+\sqrt2=6\sqrt2$
$a_7=a_6+d =6\sqrt2+\sqrt2=7\sqrt2$
Therefore, next three terms of given series are $5\sqrt2,6\sqrt2,7\sqrt2$

That is the next three terms are $\sqrt{50},\ \sqrt{72},\ \sqrt{98}$

Q4 (xiii) Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...$

Answer:

Given series is
$\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...$
Now,
the first term to this series is = $\sqrt3$
Now,
$a_1 = \sqrt3 \ \ and \ \ a_2 = \sqrt6 \ \ and \ \ a_3 = \sqrt9 \ \ and \ \ a_4 = \sqrt{12}$
$a_2-a_1 = \sqrt6-\sqrt3 =\sqrt3(\sqrt2-1)$
$a_3-a_2 = 3-\sqrt3 =\sqrt3(\sqrt3-1)$
We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xiv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small 1^2,3^2,5^2,7^2,...$

Answer:

Given series is
$\small 1^2,3^2,5^2,7^2,...$
we can rewrite it as
$1,9,25,49,....$
Now,
the first term to this series is = 1
Now,
$a_1 =1 \ \ and \ \ a_2 = 9 \ \ and \ \ a_3 =25 \ \ and \ \ a_4 = 49$
$a_2-a_1 = 9-1 = 8$
$a_3-a_2 = 25-9=16$
We can clearly see that the difference between terms are not equal
Hence, given series is not in AP

Q4 (xv) Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small 1^2,5^2,7^2,73,...$

Answer:

Given series is
$\small 1^2,5^2,7^2,73,...$
we can rewrite it as
$1,25,49,73....$
Now,
the first term to this series is = 1
Now,
$a_1 =1 \ \ and \ \ a_2 = 25 \ \ and \ \ a_3 =49 \ \ and \ \ a_4 = 73$
$a_2-a_1 = 25-1 = 24$
$a_3-a_2 = 49-25=24$
$a_4-a_3 = 73-49=24$
We can clearly see that the difference between terms are equal and equal to 24
Hence, given series is in AP
Now, the next three terms are
$a_5=a_4+d = 73+24=97$
$a_6=a_5+d = 97+24=121$
$a_7=a_6+d = 121+24=145$
Therefore, the next three terms of given series are 97,121,145