Maharashtra Board Class 10 Math Part 1 Solution Chapter 3 Practice Set 3.1 Arithmetic Progression

Maharashtra Board Class 10 Math Part 1 Solution Chapter 3 Practice Set 3.1 – Arithmetic Progression

 

Balbharati Maharashtra Board Class 10 Math Part 1 Solution Chapter 3: Arithmetic Progression. Marathi or English Medium Students of Class 10 get here Arithmetic Progression full Practice set Solution.

Std

Maharashtra Class 10
Subject

Math Part 1 Solution

Chapter

Arithmetic Progression
Practice set

3.1


Practice Set 3.1

(1) Which of the following sequences are A.P.? If they are A.P. find the common difference.

(1) 2, 4, 6, 8, ……

Solution:

2, 4, 6, 8, ……

t2 – t1 = 4 – 2 = 2; t4 – t3 = 8 – 6 = 2

t3 – t2 = 6 – 4 = 2

∴ d = 2

∴ The above series is in A.P.

 

(2) 2, 5/2, 3, 7/3 ……

Solution:

2, 5/2, 3, 7/3

t2 – t1 = 5/2 – 2

= ½

t3 – t2 = 3 – 5/2

= ½

t4 – t3 = 7/3 – 3   ∴ d = 1/2

= 1/2

∴ The above series in A.P.

 

(3) -10, -6, -2, 2, ….

Solution:

-10, -6, -2, 2, ….

t2 – t1 = -6 – (-10)

= 4

t3 – t2 = -2 – (-6)

= 4

t4 – t3 = 2 – (-1)

= 4

∴ d = 4

∴ The above series is in A.P.

 

(4) (4) 0.3, 0.33, .0333, …..

Solution:

0.3, 0.33, .0333, ……

t2 – t1 = 0.33 – 0.3

= 0.33

t3 – t2 = 0.333 – 0.33

= 0.003

∴ t2 – t1 ≠ t3 – t2

∴ The above series is not in A.P.

 

(5) 0, -4, -8, -12, ……

Solution:

0, -4, -8, -12, ……

t2 – t1 = -4 -0

= -4

t3 – t2 = – 8 – (-4)

= -4

t4 – t3 = -12 – (-8)

= -4

∴ d = -4

∴ The above series is in A.P.

 

(6) – 1/5, – 1/5, – 1/5, ……

Solution:

– 1/5, – 1/5, – 1/5, ……

Since all the terms are same

∴ d = 0

∴ The above series is in A.P.

 

 

(7) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ….

Solution:

3, 3 + √2, 3 + 2√2, 3 + 3√2, ….

t2 – t1 = 3 + √2 – 3

= √2

t3 – t2 = 3 + 2√2 – 3 – √2

= √2 (2 – 1)

= √2 × 1

= √2

t4 – t3 = 3 + 3√2 – 3 – 2√2

= √2 (3 – 2)

= √2 × 1

= √2

∴ d = √2

∴ The above series is in A.P

 

 

(8) 127, 132, 137, ……

Solution:

127, 132, 137, ……

t2 – t1 = 132 – 127

= 5

t3 – t2 = 137 – 132

∴ d = 5

∴ The above series is in A.P.

 

(2) Write an A.P. whose first term is a and common difference is d in each of the following.

(1) a = 10, d = 5

Solution:

a = 10, d = 5

∴ t2 – t1 = 5

Or, t2 – 10 = 5

Or, t2 = 15

Similarly, t3 = 20

t4 = 025

∴ The A.P: 10, 15, 20, 25 …..

 

(2) a = -3, d = 0

Solution:

a = -3, d = 0

Since d = 0

∴ The AP is -3, -3, -3, …..

 

(3) a = -7, d = ½

Solution:

a = -7, d = ½

∴ t2 – t1 = 1/2

Or, t2 – (-7) = 1/2

Or, t2 = 1/2 – 7

= 1-14/2

= -13/2

 

t3 – t2 = ½

Or, t3 – (- 13/2) = 1/2

Or, t3 = 1/2 – 13/2

= 1-13/2

= -12/2 = – 6

 

t4 – t3 = ½

Or, t4 – (-6) = 1/2

Or, t4 = 1/2 – 6

= 1-12/2 = -11/2

∴ The AP is -7, -13/2, -6, -11/2

 

 

(4) a = -1.25, d = 3

Solution:

a = -1.25, d = 3

∴ t2 – t1 = 3

Or, t2 – (1.25) = 3

Or, t2 = 3 – 1.25

= 1.75

 

t3 – t2 = 3

Or, t3 – 1.75 = 3

Or, t3 = 3 + 1.75

= 4.75

 

t4 – t3 = 3

Or, t4 – 4.75 = 3

Or, t4 = 7.75

∴ The AP is – 1.25, 1.75, 4.75, 7.75, ……

 

(5) a = 6, d = -3

Solution:

a = 6, d = -3

∴ t2 – t1 = -3

Or, t2 – 6 = -3

Or, t2 = -3

 

t3 – t2 = -3

Or, t3 – 3 = -3

Or, t3 = 0

 

t4 – t3 = -3

Or, t4 – 0 = -3

Or, t4 = -3

∴ The AP is 6, 3, 0, -3, ….

 

(6) a = -19, d = -4

Solution:

a = -19, d = -4

t2 – t1 = -4

Or, t2 – (-19) = -4

Or, t2 = -4 – 19

= -23

 

t3 – t2 = -4

Or, t3 – (-23) = -4

Or, t3 = -4 -23

= -27

 

t4 – t3 = -4

Or, t4 – (-27) = -4

Or, t4 = -4 -27 = -31

∴ The AP is -19, -23, -27, -31, …..

 

(3) Find the first term and common difference for each of the A.P.

(1) 5, 1, -3, -7, ……

Solution:

5, 1, -3, -7, ….

a = 5, t2 – t1 = 1 – 5 = -4

t3 – t2 = -3 -1 = -4

t4 – t3 = -7 – (-3) = -7 + 3

= -4

∴ d = -4,

 

(2) 0.6, 0.9, 1.2, 1.5, ……

Solution:

0.6, 0.9, 1.2, 1.5, ……

a = 0.6; t2 – t1 = 0.9 – 0.6

= 0.3

t3 – t2 = 1.2 – 0.9

= 0.3

b = t4 – t3 = 1.5 – 1.2

= 0.3

∴ d = 0.3

 

(3) 127, 135, 143, 151, …..

Solution:

127, 135, 143, 151, …..

a = 127;

t2 – t1 = 135 – 127 = 8

t3 – t2 = 143 – 135 = 8

t4 – t3 = 151 – 143 = 8

∴ d = 8

 

(4) 1/4, ¾, 5/4, 7/4, ….

Solution:

1/4, ¾, 5/4, 7/4, ….

a = 1/4;

t2 – t1 = ¾ – ¼ = 2/4 = ½

t3 – t2 = 5/4 – 3/4 = 2/4 = 1/2

t4 – t3 = 7/4 – 5/4 = 2/4 = 1/2

∴ d = ½

Updated: November 15, 2021 — 2:33 pm

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