Find the smallest number by which 18,816 must be divided do that the quotient is a perfect square. Find the square root of the quotient.
Answers
Step-by-step explanation:
NCERT Exemplar Class 10 Maths Chapter 5 Arithmetic Progressions
June 28, 2019 by Bhagya
NCERT Exemplar Class 10 Maths Chapter 5 Arithmetic Progressions
NCERT Exemplar Class 10 Maths Chapter 5 Exercise 5.1
Choose the correct answers from the given four options:
Question 1.
In an AP, if d = -4, n = 7, an = 4, then a is equal to
(A) 6
(B) 7
(C) 20
(D) 28
Solution:
(D)
In an AP,
an = a + (n – 1 )d
⇒ 4 = a + (7 – 1) (-4) [by given conditions]
⇒ 4 = a + 6(-4) ⇒ 4 + 24 = a
∴ a = 28
Question 2.
In an AP, if a = 3.5, d = 0,n = 101, then an will be
(A) 0
(B) 3.5
(C) 103.5
(D) 104.5
Solution:
(B)
In an AP,
an = a + (n – 1)d = 3.5 + (101 – 1) × 0
[by given conditions]
∴. an = 3.5
Question 3.
The list of numbers – 10, – 6, – 2, 2,… is
(A) an AP with d = -16
(B) anAP with d = 4
(C) an AP with d = – 4
(D) not an AP
Solution:
(B)
The given numbers are -10, -6, -2, 2,
Here, a1 = -10, a2 = -6, a3 = 2 and a4 = 2,….
Since a2 – a1 = -6 – (-10) = -6 + 10 = 4
a3 – a2 = -2 – (-6) = -2 + 6 = 4
a4 – a3 = 2-(-2) = 2 + 2 = 4
………….
…………
………….
Each successive term of given list has same difference i.e., 4
So, the given list forms an AP with common difference, d = 4
Question 4.
The 11th term of the AP: -5, −52 , 0, 52,…
(A) -20
(B) 20
(C) -30
(D) 30
Solution:
(B)
Given AP -5, −52 , 0, 52,…
Here, a = -5, d = −52+5=52
∴ an = a + (11 – l)d [∵ an = a + (n – 1)d]
= -5 + (10) × 52
= -5 + 25 = 20
Question 5.
The first four terms of an AP, whose first term is -2 and the common difference is -2, are
(A) -2, 0, 2, 4
(B) -2, 4, -8, 16
(C) -2, -4, -6, -8
(D) -2, -4, -8, -16
Solution:
(C)
Let the first four terms of an AP are
a, a + d, a + 2d and a + 3d
Given, that first term, a = -2 and common
difference, d = -2, then we have an AP as follows
-2, -2 -2, -2 + 2(-2), -2 + 3(-2)
= -2, -4, -6, -8
Question 6.
The 21st term of the AP whose first two terms are -3 and 4 is
(A) 17
(B) 137
(C) 143
(D) -143
Solution:
(B)
Given, first two terms of an AP are a = -3 and a + d = 4
⇒ — 3 + d = 4
Common difference, d = 7
⇒ a21 = a + (21 – 1)d [∵ an = a + (n – 1)d]
= -3 +(20)7
= -3 + 140 = 137
Question 7.
If the 2nd term ofanAPis13and the 5th term is 25, what is its 7th term?
(A) 30
(B) 33
(C) 37
(D) 38
Solution:
(B)
Given a2 = 13 and a5 = 25
⇒ a + (2 – 1)d = 13 [∵ an = a + (n – 1)d]
and a + (5 – 1 )d = 25
⇒ a + d = 13 ……….. (i)
and a + 4d = 25 ………… (ii)
On subtracting Eq. (i) from Eq. (ii), we get
3d = 25 – 13 = 12
⇒ d = 4
From Eq. (i) a = 13 – 4 = 9
∴ a7 = a + (7 – 1)d = 9 + 6 × 4 = 33
Question 8.
Which term of the AP: 21, 42, 63, 84,… is 210?
(A) 9th
(B) 10th
(C) 11th
(D) 12th
Solution:
(B)
Let nth term of the given AP be 210.
Here, first term, a = 21
and common difference, d = 42 – 21 = 21 and an = 210
∵ an = a + (n – 1 )d
⇒ 210 = 21 + (n- 1)21
⇒ 210 = 21 + 21n- 21
⇒ 210 = 21n ⇒ n = 10
Hence, the 10th term of an AP is 210.
Question 9.
If the common difference of an AP is 5, then what is a18 – a13?
(A) 5
(B) 20
(C) 25
(D) 30
Solution:
(C)
Given, the common difference of AP i.e., d = 5
Now, a18 – a13 = a + (18 – 1)d – [a + (13 – 1)d]
[∵ an = a + (n – 1)d] = a + (17 × 5) – a – (12 × 5)
= 85 – 60 = 25
Question 10.
What is the common difference of an AP in which a18 – a14 = 32?
(A) 8
(B) -8
(C) -4
(D) 4
Solution
(A)
Given, a18 – a14 = 32
⇒ a + (18 – 1)d – [a + (14 – 1)d] = 32
[∵an = a + (n – 1 )d]
⇒ a + 17d – a – 13d = 32
⇒ 4d = 32
∴ d = 8
which is the required common difference of an AP.
Question 11.
Two APs have the same common difference. The first term of one of these is -1 and that of the other is – 8. Then the difference between their 4th terms is
(A) -1
(B) -8
(C) 7
(D) -9
Solution:
(C)
Let the common difference of two APs are d1 and d2, respectively.
By given condition, d1 = d2 = d ……….. (i)
Let the first term of first AP(a1) = -1
and the first term of second AP (a2) = -8
We know that, the nth term of an AP
Tn = a + (n – 1)d