R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions
Chapter 11 - Arithmetic Progressions Ex. 5B
Let the required numbers be (a - d), a, (a + d)
Sum of these numbers = (a - d) + a + (a + d) = 3a
sum of these squares =
Sum of three numbers = 21, sum of squares of these numbers = 165
3a = 21
a = 7
Thus, a = 7 and d =
Hence, the required numbers are (4, 7, 10) or (10, 7, 4)
If are consecutive terms of an AP, then
Let the required numbers be (a - d), a and (a + d)
Sum of these numbers = (a - d) + a + (a + d) = 3a
Product of these numbers = (a - d) × a × (a + d) =
But sum = 15 and product = 80
Hence, the required numbers are (2, 5, 8)
Let the required numbers be (a - d), a, (a + d)
Sum of these number = (a - d) + a + (a + d) = 3a
Product of these numbers = (a - d) a (a + d)
But,sum = 3 and product = - 35
Thus, a = 1 and d = 6
Hence, the required numbers are (-5, 1, 7)
Let the required number be (a - d), a and (a + d)
Sum of these numbers = (a - d) + a + (a + d) = 3a
Product of these numbers = (a - d) x a x (a + d)
But sum = 24 and product = 440
Thus, a = 8 and d = 3
Hence the required numbers are (5, 8, 11)
Let the required angles be (a - 3d)°, (a - d) °, (a + d) ° and (a + 3d) °
Common difference = (a - d) - (a- 3d) = a - d - a + 3d = 2d
Common difference = 10°
2d = 10° = d = 5°
Sum of four angles of quadrilateral = 360°
First angle = (a - 3d)° = (90 - 3 × 5) ° = 75°
Second angle = (a - d)° = (90 - 5) ° = 85°
Third angle = (a + d)° = (90 + 5°) = 95°
Fourth angle = (a + 3d)° = (90 + 3 × 5)° = 105°
Let the required number be (a - 3d), (a - d), (a + d) and (a + 3d)
Sum of these numbers = (a - 3d) + (a - d)+ (a + d) + (a + 3d)
4a = 28
a = 7
Sum of the squares of these numbers
Hence, the required numbers (4, 6, 8, 10)
Chapter 11 - Arithmetic Progressions Ex. 5C
It is given that -----(1)
Now, 20th term
=(sum of first 20 term) - (sum of first 19 terms)
Putting = 20 in (1) we get
Putting n= 19 in (1), we get
Hence, the 20thterm is 99
(i)The nth term is given by
(ii)Putting n = 1 in (1) , we get
(iii)Putting n = 2 in (1), we get = 8
Here a = 21, d = (18 - 21) = -3
Let the required number of terms be n, then
sum of first 15 terms = 0
The term AP is
All numbers between 300 and 700 that are multiples of 9 are 306, 315, 324, 333, ..., 693
This is an AP in which a = 306, d = (315 - 306) = 9, l = 693
Let the number of these terms be n, then
Thus, common difference = 3
The given AP is
Common difference d =
Term next to
The given AP is
First term
Common difference d =
Sum of n terms =
First term 'a' of an AP = 2
The last term l = 29
common difference = 3
First term of an AP, a = 22
Last term = nth term = - 11
Thus, n = 12, d = -3
Chapter 11 - Arithmetic Progressions MCQ
Chapter 5 - Arithmetic Progressions Ex. 5A
The given AP is
First term = 6, common difference =
a = 6, d =
The nth term is given by
Hence, 37th term is 69
The given AP is
The first term = 5,
common difference =
a = 5,
The nth term is given by
Hence the 25th term is - 7
(i)First term = -6
(ii)Common difference
(iii)16th term = where a = -6 and d = 4
= (-6 + 15 4) = 54
In the given AP, we have a = 6 and d = (10 - 6) = 4
Suppose there are n terms in the given AP, then
Hence there are 43 terms in the given AP
In the given AP we have a = 41 and d = 38 - 41 = - 3
Suppose there are n terms in AP, then
Hence there are 12 terms in the given AP
In the given AP, we have a = 3 and d = 8 - 3 = 5
Suppose there are n terms in given AP, then
Hence, the 18th term of given AP is 88
In the given AP, we have a = 72 and d = 68 - 72 = - 4
Suppose there are n terms in given AP, we have
Hence, the 19th term in the given AP is 0
In the given AP, we have
Suppose there are n terms in given AP, we have
Then,
Thus, 14th term in the given AP is 3
The given AP is 5, 15, 25....
a = 5, d = 15 - 5 = 10
Thus, the required term is 44th
In the given AP let the first term = a,
And common difference = d
So the required AP is 7, 12, 17, 22....
Here a = 7, d = (10 - 7) = 3, l = 184
And n = 8
Hence, the 8th term from the end is 163
Here a = 17, d = (14 - 17) = -3, l = -40
And n = 6
Now, nth term from the end = [l - (n - 1)d]
Hence, the 6th term from the end is - 25
In the given AP, let the first term = a common difference = d
So the required AP is 8, 6, 4, 2, 0......
Let the first term of given AP = a and common difference = d
Hence 25th term is triple its 11th term
Let a be the first term and d be the common difference
First AP is 63, 65, 67....
First term = 63, common difference = 65 - 63 = 2
nthterm = 63 + (n - 1) 2 = 63 + 2n - 2 = 2n + 61
Second AP is 3, 10, 17 ....
First term = 3, common difference = 10 - 3 = 7
nth term = 3 + (n - 1) 7 = 3 + 7n - 7 = 7n - 4
The two nth terms are equal
2n + 61 = 7n - 4 or 5n = 61 + 4 = 65
Let a be the first term and d be the common difference
pth term = a +(p - 1)d = q(given)-----(1)
qth term = a +(q - 1) d = p(given)-----(2)
subtracting (2) from (1)
(p - q)d = q - p
(p - q)d = -(p - q)
d = -1
Putting d = -1 in (1)
Let a be the first term and d be the common difference
nth term from the beginning = a + (n - 1)d-----(1)
nth term from end= l - (n - 1)d ----(2)
adding (1) and (2),
sum of the nth term from the beginning and nth term from the end = [a + (n - 1)d] + [l - (n - 1)d] = a + l
Number of rose plants in first, second, third rows.... are 43, 41, 39... respectively.
There are 11 rose plants in the last row
So, it is an AP . viz. 43, 41, 39 .... 11
a = 43, d = 41 - 43 = -2, l = 11
Let nth term be the last term
Hence, there are 17 rows in the flower bed.
Chapter 5 - Arithmetic Progressions Ex. 5D
All three digit natural numbers divisible by 13 are 104, 117, 130, 143,..., 988
This is an AP in which a = 104, d = (117 - 104) = 13, l = 988
First 15 multiples of 8 are 8, 16, 24, ... to 15th term
Odd natural numbers between 0 and 50 are 1, 3, 5, ... 49
a = 1, d = 3 - 1= 2, l = 49
Let the number of terms be n
First 100 even natural numbers divisible by 5 are
10, 20, 30, ... to 100 term
First term of AP = 10
Common difference d = 20 - 10 = 10
Number of terms = n = 100
The given AP is 21, 18, 15, ....
First term = 21, common difference = 18 - 21= - 3
Let nth term be zero a + (n - 1)d = 0or 21 + (n - 1)(-3) = 0
21 - 3n + 3 = 0
3n = 24
or
Hence, 8th term of given series is 0
Sum of n natural numbers = 1 + 2 + 3 + ... + n
Here a = 1, d = 2 - 1 = 1
Sum of even natural numbers = 2 + 4 + 6 + ... to n terms
a = 2, d = 4 - 2 = 2
First term of AP = a = p
Common difference = d = q
nth term = a + (n 1)d
10th term = p + (10 1)q
= p + 9q
Other Chapters for CBSE Class 10 Maths
Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- ProbabilityR S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects
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