December 4, 2021

# Class10 Maths Exercise 5.1 Q(2) Q(3) Q(4)

## Class10 Maths Exercise 5.1 – Q(2) Q(3) Q(4)

Class10 Maths Exercise 5.1 – In this Question, I have explained Arithmetic Progression Concepts. Some series are AP or not is explained in the video and Written solution of Chapter 5 Exercise 5.1 Q(2) .

CLASS 10 MATHEMATICS

CH-5 ARITHMETIC PROGRESSION (AP)

EX-5.1

Q2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:

(i) a = 10, d = 10

a = 10, d = 10

Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

THUS,

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = – 2, d = 0

Let the AP be a1, a2, a3, a4, a5 …

a1 = a = -2

a2 = a1+d = – 2+0 = – 2

a3 = a2+d = – 2+0 = – 2

a4 = a3+d = – 2+0 = – 2

Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …

THUS,

First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) a = 4, d = – 3

Let the AP be  a1, a2, a3, a4, a5 …

a1 = a = 4

a2 = a1+d = 4-3 = 1

a3 = a2+d = 1-3 = – 2

a4 = a3+d = -2-3 = – 5

Therefore, the A.P. series will be 4, 1, – 2 – 5 …

THUS,

first four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) a = – 1, d = 1/2

Let the AP be a1, a2, a3, a4, a5 …

a2 = a1+d = -1+1/2 = -1/2

a3 = a2+d = -1/2+1/2 = 0

a4 = a3+d = 0+1/2 = 1/2

Therefore, the A.P. series will be-1, -1/2, 0, 1/2

THUS, First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25

:

Let the AP be a1, a2, a3, a4, a5 …

a1 = a = – 1.25

a2 = a1 + d = – 1.25-0.25 = – 1.50

a3 = a2 + d = – 1.50-0.25 = – 1.75

a4 = a3 + d = – 1.75-0.25 = – 2.00

Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

THUS, first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

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CLASS 10 MATHEMATICS

CH-5 ARITHMETIC PROGRESSION (AP)

EX-5.1

Q3. For the following A.P.s, write the first term and the common difference.

(i) 3, 1, – 1, – 3 …

Given series;

3, 1, – 1, – 3 …

First term, a = 3

Common difference, d = a2-a1 or

an – an-1

therefore,

d=1 – 3 = -2

THUS, d = -2

(ii) Given series, – 5, – 1, 3, 7 …

First term, a = -5

Common difference, d = a2-a1

therefore,

d= ( – 1)-( – 5) = – 1+5 = 4

THUS, d=4

(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….

First term, a = 1/3

Common difference, d = a2-a1

therefore,

d= 5/3 – 1/3 = 4/3

THUS,

d=4/3

(iv) Given series, 0.6, 1.7, 2.8, 3.9 …

First term, a = 0.6

Common difference, d = a2-a1

therefore,

d=1.7 – 0.6=1.1

THUS,

d=1.1

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CLASS 10 MATHEMATICS

CH-5 ARITHMETIC PROGRESSION (AP)

EX-5.1

Q4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.

(i) 2, 4, 8, 16 …

given,

2, 4, 8, 16 …

Here, the common difference is;

a2 – a1 = 4 – 2 = 2

a3 – a2 = 8 – 4 = 4

a4 – a3 = 16 – 8 = 8

Since, an+1 – an or the common difference is not the same every time.

Therefore, the given series are not forming an A.P.

(ii) 2, 5/2, 3, 7/2 ….

Given, 2, 5/2, 3, 7/2 ….

Here,

a2 – a1 = 5/2-2 = 1/2

a3 – a2 = 3-5/2 = 1/2

a4 – a3 = 7/2-3 = 1/2

Since, an+1 – an or the common difference is same every time.

Therefore, d = 1/2 and the given series are in A.P.

The next three terms are;

a5 = 7/2+1/2 = 4

a6 = 4 +1/2 = 9/2

a7 = 9/2 +1/2 = 5

(iii) -1.2, -3.2, -5.2, -7.2 …

Given

, -1.2, – 3.2, -5.2, -7.2 …

Here,

a2 – a1 = (-3.2)-(-1.2) = -2

a3 – a2 = (-5.2)-(-3.2) = -2

a4 – a3 = (-7.2)-(-5.2) = -2

Since, an+1 – an or common difference is same every time.

Therefore, d = -2 and the given series are in A.P.

Hence, next three terms are;

a5 = – 7.2-2 = -9.2

a6 = – 9.2-2 = – 11.2

a7 = – 11.2-2 = – 13.2

(iv) -10, – 6, – 2, 2 …

given,

-10,-6,-2,2

Here, the terms and their difference are;

a2 – a1 = (-6)-(-10) = 4

a3 – a2 = (-2)-(-6) = 4

a4 – a3 = (2 -(-2) = 4

Since, an+1 – an or the common difference is same every time.

Therefore, d = 4 and the given numbers are in A.P.

Hence, next three terms are;

a5 = 2+4 = 6

a6 = 6+4 = 10

a7 = 10+4 = 14

(v) 3, 3+√2, 3+2√2, 3+3√2

GIVEN,

3, 3+√2, 3+2√2, 3+3√2

Here,

a2 – a1 = 3+√2-3 = √2

a3 – a2 = (3+2√2)-(3+√2) = √2

a4 – a3 = (3+3√2) – (3+2√2) = √2

Since, an+1 – an or the common difference is same every time.

Therefore, d = √2 and the given series forms a A.P.

Hence, next three terms are;

a5 = (3+√2) +√2 = 3+4√2

a6 = (3+4√2)+√2 = 3+5√2

a7 = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

GIVEN,

3, 3+√2, 3+2√2, 3+3√2

Here,

a2 – a1 = 0.22-0.2 = 0.02

a3 – a2 = 0.222-0.22 = 0.002

a4 – a3 = 0.2222-0.222 = 0.0002

Since, an+1 – an or the common difference is not same every time.

Therefore, and the given series doesn’t forms a A.P.

(vii) 0, -4, -8, -12 …

GIVEN,0, -4, -8, -12 …

Here,

a2 – a1 = (-4)-0 = -4

a3 – a2 = (-8)-(-4) = -4

a4 – a3 = (-12)-(-8) = -4

Since, an+1 – an or the common difference is same every time.

Therefore, d = -4 and the given series forms a A.P.

Hence, next three terms are;

a5 = -12-4 = -16

a6 = -16-4 = -20

a7 = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

GIVEN, -1/2, -1/2, -1/2, -1/2 ….

Here,

a2 – a1 = (-1/2) – (-1/2) = 0

a3 – a2 = (-1/2) – (-1/2) = 0

a4 – a3 = (-1/2) – (-1/2) = 0

Since, an+1 – an or the common difference is same every time.

Therefore, d = 0 and the given series forms a A.P.

Hence, next three terms are;

a5 = (-1/2)-0 = -1/2

a6 = (-1/2)-0 = -1/2

a7 = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

GIVEN,1, 3, 9, 27 …

Here,

a2 – a1 = 3-1 = 2

a3 – a2 = 9-3 = 6

a4 – a3 = 27-9 = 18

Since, an+1 – an or the common difference is not same every time.

Therefore, and the given series doesn’t form a A.P.

(x) a, 2a, 3a, 4a …

GIVEN,a, 2a, 3a, 4a …

Here,

a2 – a1 = 2aa

a3 – a2 = 3a-2a = a

a4 – a3 = 4a-3a = a

Since, an+1 – an or the common difference is same every time.

Therefore, d = a and the given series forms a A.P.

Hence, next three terms are;

a5 = 4a+a = 5a

a6 = 5a+a = 6a

a7 = 6a+a = 7a

(xi) aa2, a3, a4 …

GIVEN,aa2, a3, a4 …

Here,

a2 – a1 = a2–a = a(a-1)

a3 – a2 = a3 – a2 = a2(a-1)

a4 – a3 = a4 – a3 = a3(a-1)

Since, an+1 – an or the common difference is not same every time.

Therefore, the given series doesn’t forms a A.P.

(xii) √2, √8, √18, √32 …

GIVEN,√2, √8, √18, √32 …

Here,

a2 – a1 = √8-√2  = 2√2-√2 = √2

a3 – a2 = √18-√8 = 3√2-2√2 = √2

a4 – a3 = 4√2-3√2 = √2

Since, an+1 – an or the common difference is same every time.

Therefore, d = √2 and the given series forms a A.P.

Hence, next three terms are;

a5 = √32+√2 = 4√2+√2 = 5√2 = √50

a6 = 5√2+√2 = 6√2 = √72

a7 = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

GIVEN,√3, √6, √9, √12 …

Here,

a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)

a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)

a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

Since, an+1 – an or the common difference is not same every time.

Therefore, the given series doesn’t form a A.P.

(xiv) 12, 32, 52, 72 …

GIVEN,12, 32, 52, 72 …

Here,

a2 − a1 = 32−12 = 20

a3 − a2 = 52−32 = 20

a4 − a3 = 72−52= 20

Since, an – an-1  or the common difference is same every time.

Therefore, the given series forms a A.P.

The next three terms are

a5=a4+d=92

a6=a5+d=112

a7=a6+d=132

12,32,52,72,92, 112 ,132 are in AP

(xv) 12, 52, 72, 73 …

Given, 12, 52,72,73….

Here,

a2 − a1 = 52−12 = 40

a3 − a2 = 72−52= 20

a4 − a3 = 73−72 = 1

Since, anan-1 or the common difference isn’t same every time.

Therefore, the series isn’t AP