arithmetic progression important question and answers class 10
Answers
Answer:
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1. Second example: the sequence 3, 5, 7, 9, 11,... is an arithmetic progression. with common difference 2
Step-by-step explanation:
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Answer:
Question 1.
Find the common difference of the AP 1p,1−pp,1−2pp,…. (2013D)
Solution:
The common difference,

Question 2.
Find the common difference of the A.P. 12b,1−6b2b,1−12b2b,…. (2013D)
Solution:
The common difference, d = a2 – a1 = 1−6b2b−12b
= 1−6b−12b=−6b2b = -3
Question 3.
Find the common difference of the A.P. 13q,1−6q3q,1−12q3q,…. (2013D)
Solution:
Common difference, d = a2 – a1 = 1−6q3q−13q
= 1−6q−13q=−6q3q = -2
Question 4.
Calculate the common difference of the A.P. 1b,3−b3b,3−2b3b,…. (2013D)
Solution:
Common difference, d = a2 – a1= 3−b
Question 3.
Find the common difference of the A.P. 13q,1−6q3q,1−12q3q,…. (2013D)
Solution:
Common difference, d = a2 – a1 = 1−6q3q−13q
= 1−6q−13q=−6q3q = -2
Question 4.
Calculate the common difference of the A.P. 1b,3−b3b,3−2b3b,…. (2013D)
Solution:
Common difference, d = a2 – a1= 3−b3b−1b
= 3−b−33b=−b3b=−13
Question 5.
Calculate the common difference of the A.P. 13,1−3b3,1−6b3,… (2013OD)
Solution:
Common difference, d = a2 – a1 = 1−3b3−13
= 1−3b−13=−3b3 = -b
Question 6.
What is the common difference of an A.P. in which a21 – a7 = 84? (2017OD )
Solution:
a21 – a7 = 84 …[Given
∴ (a + 20d) – (a + 6d) = 84 …[an = a + (n – 1)d
20d – 6d = 84
14d = 84 ⇒ d 8414 = 6
Question 7.
Find the 9th term from the end (towards the first term) of the A.P. 5,9,13, …, 185. (2016D)
Solution:
Here First term, a = 5
Common difference, d = 9 – 5 = 4
Last term, 1 = 185
nth term from the end = l – (n – 1)d
9th term from the end = 185 – (9 – 1)4
= 185 – 8 × 4 = 185 – 32 = 153
Question 8.
The angles of a triangle are in A.P., the least being half the greatest. Find the angles. (2011D)
Solution:
Let the angles be a – d, a, a + d; a > 0, d > 0
∵ Sum of angles = 180°
∴ a – d + a + a + d = 180°
⇒ 3a = 180° ∴ a = 60° …(i)
By the given condition
a – d = a+d2
⇒ 2 = 2a – 2d = a + d
⇒ 2a – a = d + 2d ⇒ a = 3d
⇒ d = a3=60∘3 = 20° … [From (i)
∴ Angles are: 60° – 20°, 60°, 60° + 20°
i.e., 40°, 60°, 80°
Question 9.
Find whether -150 is a term of the A.P. 17, 12, 7, 2, … ? (2011D)
Solution:
Given: 1st term, a = 17
Common difference, d = 12 – 17 = -5
nth term, an = – 150 (Let)
∴ a + (n – 1) d = – 150
17 + (n – 1)(-5) = – 150
(n – 1) (-5) = – 150 – 17 = – 167
(n − 1) = −167−5
n = 1675 + 1 = 167+55=1725
n = 1725 …[Being not a natural number
∴ -150 is not a term of given A.P.
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