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