# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression Exercise 9.3

**1. For the following arithmetic progressions write the first term a and the common difference d :**

**(i) -5, -1, 3, 7, ……(ii) 1/5 , 3/5 , 5/5 , 7/5 , ……(iii) 0.3, 0.55, 0.80, 1.05, ……(iv) -1.1, -3.1, -5.1, -7.1, ……**

**Solution**

Here, first term (a) = -5

We have, a

_{2}- a

_{1}= -1 - (-5) = 4

a

_{3}- a

_{2}= 3 - (-1) = 3 + 1 = 4

a

_{4}- a

_{3}= 7 - 3 = 4

∴ Common difference (d) = 4

(iii) 0.3, 0.55, 0.80, 1.05,....

Here, first term (a) = 0.3

We have = a_{2}- a_{1} = 0.55 - 0.3 = 0.25

a_{3} - a_{2} = 0.80 - 0.55 = 0.25

∴ Common difference = 0.25

(iv) -1.1, -3.1, -5.1, -7.1,....

Here, first term (a) = -1.1

We have = a_{2} - a_{1} = -3.1 - (-1.1)

= -3.1 + 1.1 = -2.0

a_{3} - a_{2} = -5.1 - (-3.1) = -5.1 + 3.1 = -2.0

a_{4} - a_{3} = -7.1 - (-5.1) = -7.1 + 5.1 = -2.0

∴ Common difference = -2.0

**2. Write the arithmetic progression when first term a and common difference d are as follows:**

**(i) a = 4, d = -3(ii) a = -1, d= 12(iii) a = -1.5, d = -0.5**

**Solution**

(i) First term (a) = 4

and common difference (d) = -3

Second term = a + d = 4 – 3 = 1

Third term = a + 2d = 4 + 2×(-3) = 4 – 6 = -2

Fourth term = a + 3d = 4 + 3 (-3) = 4 – 9 = -5

Fifth term = a + 4d = 4 + 4 (-3) = 4 – 12 = -8

AP will be 4, 1, -2, -5, -8, ….

(iii) First term (a) = -1.5 and common difference (d) = -0.5

Second term = a + d = -1.5 + (-0.5)

= -1.5 - 0.5 = -2.0

Third term = a + 2d = -1.5 + 2×(-0.5)

= -1.5 - 1.0 = -2.5

Fourth term = a + 3d = -1.5 + 3×(-0.5)

= 1.5 - 1.5 = -3.0

Fifth term = a + 4d = -1.5 + 4×(-0.5)

= -1.5 - 2.0 = -3.5

AP will be -1.5, -2.0, -2.5, -3.0, -3.5,....

**3. In which of the following situations, the sequence of numbers formed will form an A.P?(i) The cost of digging a well for the first metre is ₹ 150 and rises by ₹ 20 for each succeeding metre.(ii) The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of the remaining in the cylinder.(iii) Divya deposited ₹ 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …, and so on. [NCERT Exemplar]**

**Solution**

(i) Cost of digging a well for the first metre = ₹ 150

Cost for the second metre = ₹ 150 + ₹ 20 = ₹ 170

Cost for the third metre = ₹ 170 + ₹ 20 = ₹ 190

Cost for the fourth metre = ₹ 190 + ₹ 20 = ₹ 210

The sequence will be (In rupees)

150, 170, 190, 210, ……

Which is an A.P.

Whose = 150 and d = 20

(ii) Let air present in the cylinder = 1

(iii) Amount at the end of the 1st year = ₹ 1100

Amount at the end of the 2nd year = ₹ 1210

Amount at the end of 3rd year = ₹ 1331 and so on.

So, the amount (in ₹) at the end of 1st year, 2nd year, 3rd year, … are

1100, 1210, 1331, …….

Here, a_{2} – a_{1} = 110

a_{3} – a_{2} = 121

As, a_{2} – a_{1} ≠ a_{3} – a_{2}, it does not form an A.P.

**4. Find the common difference and write the next four terms of each of the following arithmetic progressions :**

**(i) 1, -2, -5, -8, ……(ii) 0, -3, -6, -9, ….(iii) -1, 1/4 , 3/2 , ….(iv) -1, – 5/6 , – 2/3 , ….**

**Solution**

(i) 1, -2, -5, -8,....

Here, a_{1} = 1, a_{2} = -2, a_{3} = -5, a_{4} = -8,....

Now a_{2} - a_{1} = -2-1 = -3

a_{3} - a_{2} = -5-(-2) = -5 + 2 = -3

a_{4} - a_{3} = -8-(-5) = -8 + 5 = -3

It is an A.P. whose common difference is = -3

Now, next four terms will be

-8 - 3 = -11

-11 - 3 = -14

-14 - 3 = -17

-17 - 3 = -20

∴ -11, -14, -17, -20 are four term next to these.

(ii) 0, -3, -6, -9,....

Here, a_{1} = 0, a_{2} = -3, a_{3}= -6, a_{4} = -9

Now, a_{2} - a_{1} = -3 - 0 = -3

a_{3} - a_{2} = -6-(-3) = -6 + 3 = -3

a_{4} - a_{3} = -9-(-6) = -9 + 6 = -3

We see that it is an A.P. whose common difference is -3.

Now, next four terms will be

-9 - 3 = -12

-12 - 3 = -15

-15 - 3 = -18

-18 - 3 = -21

∴ -12, -15, -18, -21 are next four terms.

5. Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference ?

**Solution**

a_{n} = a + nb

Let n= 1, 2, 3, 4, 5, ……….

a_{1} = a + b

a_{2} = a + 2b

a_{3} = a + 3b

a_{4} = a + 4b

a_{5} = a + 5b

We see that it is an A.P. whose common difference is b and a for any real value of a and b as

a_{2} – a_{1} = a + 2b – a – b = b

a_{3} – a_{2} = a + 3b – a – 2b = b

a_{4} – a_{3} = a + 4b – a – 3b = b

and a_{5} – a_{4} = a + 5b – a – 4b = b

**6. Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.**

**(i) 3, 6, 12, 24,....(ii) 0, -4, -8, -12,....(iii) 1/2, 1/4, 1/6, 1/8,....(iv) 12, 2, -8, -18,....(v) 3, 3, 3, 3,....(vi) p, p + 90, p + 180, p + 270,.... where p = (999)**

^{999}(vii) 1.0, 1.7, 2.4, 3.1,....(viii) -225, -425, -625, -825,....(ix) 10, 10 + 2

^{5}, 10 + 2

^{6}, 10 + 2

^{7},....(x) a+b, (a+ 1)+b, (a+ 1)+(b+1), (a+ 2) + (b + 1), (a +2) + (b + 2),....(xi) 1

^{2}, 32, 5

^{2}, 7

^{2},....(xii) 1

^{2}, 5

^{2}, 7

^{2}, 73,....

**Solution**

(i) 3, 6, 12, 24,....

Let a_{1} = 3, a_{2} = 6, a_{3} = 12, a_{4} = 24

Now, a_{2} - a_{1} = 6 - 3 = 3

a_{3} - a_{2} = 12 - 6 = 6

a_{4} - a_{3} = 24- 12 = 12

It has no common difference.

∴ It is not an A.P.

(ii) 0, -4, -8, -12,....

Let a_{1} = 0, a_{2} = -4, a_{3} = -8, a_{4} = -12

Now, 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

Here, we see that common difference is -4.

∴ It is an A.P.

We see that there is no common difference.

It is not an A.P.

(iv) 12, 2, -8, -18,....

Let a_{1} = 12, a_{2} = 2, a_{3} = -8, a_{4} = -18

Now, a_{2} - a_{1} = 2 - 12 = -10

a_{3} - a_{2} = -8 - 2 = -10

a_{4} - a_{3} = -18 -(-8) = -18 + 8 = -10

We see that common difference is -10.

∴ It is an A.P.

(v) 3, 3, 3, 3,....

Let a_{1} = 3, a_{2} = 3, a_{3} = 3, a_{4} = 3,....

Now, a_{2} - a_{1} = 3 - 3 = 0

a_{3} - a_{2} = 3 - 3 = 0

a_{4} - a_{3} = 3 - 3 = 0

We see that common difference is 0.

∴ It is an A.P.

(vi) p, p + 90, p + 180, p + 270,.... where p = (999)^{999}

Let a_{1} = p, a_{2} = p + 90, a_{3} = p + 180, a_{4} = p + 270

Now,

a_{2} - a_{1} = p + 90 - p = 90

a_{3} - a_{2} = p + 180 - p - 90 = 90

a_{4} - a_{3} = p + 270 - p - 180 = 90

We see that common difference is 90.

∴ It is an A.P.

(vii) 1.0, 1.7, 2.4, 3.1,....

Let a_{1} = 1.0, a_{2} = 1.7, a_{3} = 2.4, a_{4} = 3.1

Now a_{2} - a_{1} = 1.7- 1.0 = 0.7

a_{3} - a_{2} = 2.4 - 1.7 = 0.7

a_{4} - a_{3} = 3.1 - 2.4 = 0.7

We see that common difference is 0.7.

∴ It is an A.P.

(viii) -225, -425, -625, -825,....

Let a_{1} = -225, a_{2} = -425, a_{3} = -625, a_{4} = -825

Now, a_{2} - a_{1} = -425 -(-225) = -425 + 225 = -200

a_{3} - a_{2} = -625 -(-425) = -625 + 425 = -200

a_{4} - a_{3} = -825 -(-625) = -825 + 625 = -200

We see that common difference is -200.

∴ It is an A.P.

(ix) 10, 10 + 2^{5}, 10 + 2^{6}, 10 + 2^{7},....

Let a_{1} = 10, a_{2} = 10 + 2^{5}, a_{3} = 10 + 2^{6}, a_{4} = 10 + 2^{7}

Now, a_{2} - a_{1} = 10 + 2^{5} - 10 = 2^{5} = 32

a_{3} - a_{2} = 10 + 2^{6} - 10 - 2^{5} = 2^{6}- 2^{5} = 64 - 32 = 32

a_{4} - a_{3} = 10 + 2^{7} - 10 - 2^{6} = 2^{7}- 2^{6} = 128 - 64 = 64

We see that there is no common difference.

∴ It is not an A.P.

(x) a+b, (a+1)+b, (a+1)+(b+1), (a+2)+(b+1), (a+2)+(b+2),....

Let a_{1} = a+b

a_{2}= (a+1)+b

a_{3} = (a+1)+(b+1)

a_{4} = (a+2)+(b+1)

a_{5} = (a+2)+(b+2)

Now, a_{2} - a_{1} = (a+1)+b-a-b = a+1+b-a-b = 1

a_{3}- a_{2} = (a+1)+(b+1)-{(a+1)+b} = a+1+b+1-a-1-b = 1

a_{4} - a_{3} = {(a+2)+(b+1)}-{(a+1)+(b+1)) = a+2+b+1-a-1-b-1 = 1

a_{5}- a_{4} = {(a+2)+(b+2)} - {(a+2)+(b+1)} = a+2+b+2-a-2-b-1 = 1

We see that common difference is 1.

∴ It is an A.P.

(xi) 1^{2}, 3^{2}, 5^{2}, 7^{2},...

Let a_{1} = 1^{2}, a_{2} = 3^{2}, a_{3}= 5^{2}, a_{4}= 7^{2}Now, a_{2} - a_{1} = 3^{2} - 1^{2} = 9-1 = 8

a_{3} - a_{2} = 5^{2} - 3^{2} = 25 - 9 = 16

a_{4} - a_{3} = 7^{2} - 5^{2} = 49 - 25 = 24

We see that there is no common difference.

∴ It is not an A.P.

(xii) 1^{2}, 5^{2}, 7^{2}, 73,....

Let a_{1} = 1^{2}, a_{2} = 5^{2}, a_{3} = 7^{2}, a, = 73

Now, a_{2} - a_{1} = 5^{2} - 1^{2} = 25 - 1 =24

a_{3} - a_{2} = 7^{2} - 5^{2} = 49 - 25 = 24

a_{4} - a_{3} = 73 - 7^{2} = 73 - 49 = 24

We see that the common difference is 24.

∴ It is an A.P.

**7. Find the common difference of the A.P. and write the next two terms :**

**(i) 51, 59, 67, 75, …….(ii) 75, 67, 59, 51, ……(iii) 1.8, 2.0, 2.2, 2.4, ……(iv) 0, 1/4 , 1/2 , 3/4 , ……(v) 119, 136, 153, 170, ……**

**Solution**