**RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progression** Ex 5.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.1.

## Rajasthan Board RBSE Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.1

RBSE Solutions For Class 10 Maths Chapter 5 Question 1.

Find the first term and common (RBSESolutions.com) difference for the following A.P –

(i) 6, 9, 12, 15, ….

(ii) – 7, – 9, – 11, – 13

(iii) \(\frac { 3 }{ 2 } ,\frac { 1 }{ 2 } ,\frac { -1 }{ 2 } ,\frac { -3 }{ 2 } ,….\)

(iv) 1, – 2, – 5, – 8 ……

(v) -1, \(\frac { 1 }{ 4 }\), \(\frac { 3 }{ 2 }\), ….

(vi) 3, 1, -1, -3, ….

(vii) 3, -2, -7, -12, …..

Solution :

(i) Given A.P. is 6, 9, 12, 15. …

First term (a_{1}) = 6

Common difference (d) = a_{2} – a_{1}

= 9 – 6 = 3

Thus, a_{1} = 6 and d = 3

(ii) Given A.P. is -7, -9, -11, -13, ……

First term (a_{1}) = – 7

Common (RBSESolutions.com) difference (d)

= a_{2} – a_{1}

= -9 – (-7)

= -9 + 7 = -2

Thus, a_{1} = -7 and d = -2

(iii) Given A.P. is \(\frac { 3 }{ 2 } ,\frac { 1 }{ 2 } ,\frac { -1 }{ 2 } ,\frac { -3 }{ 2 } ,….\)

First term (a_{1}) = \(\frac { 3 }{ 2 }\)

Common difference (d) = a_{2} – a_{1}

= \(\frac { 1 }{ 2 }\) – \(\frac { 3 }{ 2 }\)

= \(\frac { 1-3 }{ 2 }\) = \(\frac { -2 }{ 2 }\)

= -1

Thus, a_{1} = and d = – 1

(iv) Given A.P. is 1, -2, -5, -8, …….

First term (a_{1}) = 1

Common difference (d) = a_{2} – a_{1}

= -2 – (1)

= -3

Thus, a_{1} = 1 and d = -3

(v) Given A.P. -1, \(\frac { 1 }{ 4 }\), \(\frac { 3 }{ 2 }\), …..

First term (a_{1}) = – 1

Common (RBSESolutions.com) difference (d) = a_{2} – a_{1}

= \(\frac { 1 }{ 4 }\) – (-1)

= \(\frac { 1 }{ 4 }\) + 1

= \(\frac { 1+4 }{ 4 }\) = \(\frac { 5 }{ 4 }\)

Thus, a_{1} = -1 and d = \(\frac { 5 }{ 4 }\)

(vi) Given A.P. is 3, 1, -1, -3, …..

First term (a_{1}) = 3

Common difference (d) = a_{2} – a_{1}

= 1 – 3

Thus, a_{1} = 3 and d = -2

(vii) Given A.P. is 3, -2, -7, -12 , …..

First term (a_{1}) = 3

Common difference (d) = a_{2} – a_{1}

= -2 – 3

= -5

Thus, a_{1} = 3 and d = -5

RBSE Class 10 Maths Chapter 5 Question 2.

If first term a and common (RBSESolutions.com) differenced of an AP is given as follows, then find next four terms of that series.

(i) a = -1, d = \(\frac { 1 }{ 2 }\)

(ii) a = \(\frac { 1 }{ 3 }\) , d = \(\frac { 4 }{ 3 }\)

(iii) a = 0.6 , d = 1.1

(iv) a = 4, d = -3

(v) a = 11, d = -4

(vi) a = – 1.25, d = -0.25

(vii) a = 20, d = \(\frac { -3 }{ 4 }\)

Solution :

(i) Given a = -1, d = \(\frac { 1 }{ 2 }\)

First term (a) = -1

Second term (a + d) =-1 + \(\frac { 1 }{ 2 }\) = –\(\frac { -1 }{ 2 }\)

Third term (a + 2d) = -1 + 2 × \(\frac { 1 }{ 2 }\) = o

Fourth term (a + 3d) = -1 + 3 × \(\frac { 1 }{ 2 }\)

= -1 + \(\frac { 3 }{ 2 }\)

= \(\frac { -2+3 }{ 2 }\) = \(\frac { 1 }{ 2 }\)

Thus, First four terms of A.P. are – 1, – \(\frac { 1 }{ 2 }\), 0 and \(\frac { 1 }{ 2 }\).

(ii) Given a = \(\frac { 1 }{ 3 }\), d = \(\frac { 4 }{ 3 }\)

First term (a) = \(\frac { 1 }{ 3 }\)

(iii) Given a = 0.6, d = 1.1

First term (a) = 0.6

Second (RBSESolutions.com) term (a + d) = 0.6 + 1.1 = 1.7

Third terms (a + 2d) = 0.6 + 2 × 1.1 = 2.8

Fourth term (a + 3d) = 0.6 + 3 × 1.1 = 3.9

Hence, four terms of A.P are 0.6, 1.7, 2.8 and 3.9.

(iv) Given a = 4, d = -3

First term (a) = 4

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

Hence, four terms of A.P are 4, 1, -2 and -5.

(v) Given a = 11, d = -4

First term (a) = 11

Second term (a + d) = 11 + (-4) = 11 – 4 = 7

Third terms (a + 2d) = 11 + 2 (-4) = 11 – 8 = 3

Fourth terms (a + 3d) = 11 + 3 (-4) = 11 – 12 = -1

Hence, first four terms of A.P. are 11, 7, 3 and -1.

(vi) Given a = -1.25, d = -0.25

First term (a) = -1.25

Second (a + d) = -1.25 + (-0.25)

= -1.25 – 0.25 = -1.50

Third term (a + 2d)

= -1.25 + 2 (-0.25)

= -1.25 – 0.50 = – 1.75

Fourth term (a + 3d) = -1.25 + 3(-0.25)

= -1.25 – 0.75 = – 2.00

Hence first four (RBSESolutions.com) terms of A.P. are -1.25, -1.50, -1.75 and -2.00.

(vii) Given a = 20, d = \(\frac { -3 }{ 4 }\)

First Term (a) = 20

Ex 5.1 Class 10 RBSE Question 3.

Test A.P for given series of (RBSESolutions.com) numbers. For an A.P., find its common difference and next four terms also.

(i) 2, \(\frac { 5 }{ 2 }\), 3, \(\frac { 7 }{ 2 }\) ……

(ii) \(\frac { -1 }{ 2 } ,\frac { -1 }{ 2 } ,\frac { -1 }{ 2 } ,\frac { -1 }{ 2 } ,….\)

(iii) a, a^{2}, a^{3}, a^{4}, ……

(iv) √3, √6, √9, √12, ……

(v) √2, √8, √18, √32, …..

(vi) a, 2a, 3a, 4a

(vii) 0.2, 0.22, 0.222 ……

(viii) 3, 3 + √2, 3 + 2√2 , 3 + 3√2, ….

Solution :

(i) Given series 2, \(\frac { 5 }{ 2 }\), 3, \(\frac { 7 }{ 2 }\)

Here a_{1} = 2, a_{2} = \(\frac { 5 }{ 2 }\), a_{3} = 3, a_{4} = \(\frac { 7 }{ 2 }\)

Difference between two consecutive terms (d) :

∵ Difference between two (RBSESolutions.com) consecutive is same, d = \(\frac { 1 }{ 2 }\)

∴ Given series is A.P.

Next four terms are

Hence d = \(\frac { 1 }{ 2 }\) and next four (RBSESolutions.com) terms of series are 4, \(\frac { 9 }{ 2 }\) , 5 and \(\frac { 11 }{ 2 }\)

(ii) Given series \(\frac { -1 }{ 2 } ,\frac { -1 }{ 2 } ,\frac { -1 }{ 2 } ,\frac { -1 }{ 2 } ,….\)

Here a_{1} = \(\frac { -1 }{ 2 }\) , a_{2} = \(\frac { -1 }{ 2 }\), a_{3} = \(\frac { -1 }{ 2 }\), a_{4} = \(\frac { -1 }{ 2 }\)

Difference between two consecutive terms (d) :

∵ Difference between two consecutive (RBSESolutions.com) terms is same, common difference = 0

Thus, given series is a A.P.

Fifth term a_{5} = Fourth term a_{4} + Common difference d

= – \(\frac { 1 }{ 2 }\) + 0 = –\(\frac { 1 }{ 2 }\)

Sixth term a_{6} = Fifth term a_{5} + Common differece d

= – \(\frac { 1 }{ 2 }\) + 0 = – \(\frac { 1 }{ 2 }\)

Seventh term a_{7} = sixth term a_{6} + common difference d

= – \(\frac { 1 }{ 2 }\) + 0 = –\(\frac { 1 }{ 2 }\)

Eight term a_{8} = Seventh term a_{7} + Common difference d

Hence common difference d = 0 and next four terms are –\(\frac { 1 }{ 2 }\) , –\(\frac { 1 }{ 2 }\) , –\(\frac { 1 }{ 2 }\) and –\(\frac { 1 }{ 2 }\)

(iii) Given series a, a^{2}, a^{3}, a^{4}, …

Here a_{1} = a, a_{2} = a^{2}, a_{3} = a^{3}, a_{4} = a^{4}

Difference between two (RBSESolutions.com) consecutive terms d :

a_{2} – a_{1} = a^{2} – a = a(a – 1)

a_{3} – a_{2} = a^{3} – a^{2} = a^{2}(a – 1)

∵ Difference is not same

i.e., a_{2} – a_{1} ≠ a_{3} – a_{2}

Hence, given series is not an A.P.

(iv) Given series √3, √6, √9, √12, ……

a_{1} = √3, a_{2} = √6, a_{3} = √9, a_{4} = √12

Difference between two consecutive terms d :

a_{2} – a_{1} = √6 – √3 = √3(√2 – 1) = 0.717

a_{3} – a_{2} = √9 – √6 = √3(√3 – √2) = 0.530

∵ Difference is not same

i.e., a_{2} – a_{1} ≠ a_{3} – a_{2}

Thus, Given senes is not A.P

(v) Given series is √2, √8, √18, √32, …..

Here a_{1} = √2, a_{2} √8,, a_{3} = √18, a_{4} = √32

Difference between two (RBSESolutions.com) consecutive terms d :

∵ Difference is same

Common difference d = √2 and given series is an A.P.

Then 5^{th} term a_{5} = a_{4} + d

6^{th} term a_{6} = a_{5} + d

7^{th} term a_{7} = a_{6} + d

8^{th} term a_{8} = a_{7} + d

Hence, common difference d = √2 and (RBSESolutions.com) next four terms of series are √50, √72, √98 and √128

(vi) Given series is a, 2a, 3a, 4a,…

Here a_{1} = a, a_{2} = 2a, a_{3} = 3a, a_{4} = 4a

Difference between two consecutive terms (d) :

a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

a_{4} – a_{3} = 4a – 3a = a

∵ Common difference is same

Thus, common difference d = a and given series is an A.P.

Then 5^{th} term a_{5} = a_{4} + d

= 4a + a = 5a

6^{th} term a_{6} = a_{5} + d

= 5a + a = 6a

7^{th} term a_{7} = a_{6} + d

= 6a + a = 7a

8^{th} term a_{8} = a_{7} + d

= 7a + a = 8a

Hence, common (RBSESolutions.com) difference d = a and next four terms arc 5a, 6a, 7a and 8a.

(vii) Given series is

0.2, 0.22, 0.222. 0.2222, …

Hear a_{1} = 0.2, a_{2} = 0.22, a_{3} = 0.222, a_{4} = 0.2222.

Difference of two consecutive terms (d) :

a_{2} – a_{1} = 0.22 – 0.2 = 0.02

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

a_{4} – a_{3} = 0.2222 – 0.222 = 0.0002

Difference is not same

i.e., a_{2} – a_{1} ≠ a_{3} – a_{2} ≠ a_{4} – a_{3}

Hence given series ¡s not AP.

(viii) Given series is 3, 3+√2, 3+2√2, 3+3√2, …….

ie., a_{1} = 3, a_{2} = 3+√2, a_{3} = 3+√2, a_{4} = 3+√2

Difference between two consecutive terms (d) :

∵ Difference is same.

∴ Common (RBSESolutions.com) difference d = √2, and given series is an A.P.

Then, fifth term a_{5} = a_{4} + d

= 3 + 3√2 + √2

= 3 + √2(3 + 1) = 3 + 4√2

Then, sixth term a_{6} = a_{5} + d

= 3 + 4√2 + √2

= 3 + √2(4 + 1) = 3 + 5√2

Then, seventh term a_{7} = a_{6} + d

= 3 + 5√2 + √2

= 3 + √2(5 + 1) = 3 + 6√2

Then, Eighth term a_{8} = a_{7} + d

= 3 + 6√2 + √2

= 3 + √2(6 + 1) = 3 + 7√2

Hence, common difference d = √2 and next four (RBSESolutions.com) terms of series are 3 + 4√2, 3 + 5√2, 3 + 6√2 and 3 + 7√2

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