arithmetic progressions - problem based video part 4 for class 10th maths
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Arithmetic progressions - Problem based video on Arithmetic progressions part 4 for class 10th maths. About us - Lets tute is an online learning centre. We provide quality education for all learners and 24/7 academic guidance through E-tutoring. Our Mission- Our aspiration is to be a renowned unpaid school on Web-World. Contact Us - Website - www.letstute.com YouTube - www.youtube.comTRANSCRIPT
1 Chapter : Arithmetic Progressions Website: www.letstute.com
Arithmetic Progressions
Problems based on Arithmetic Progressions
Given : Sequence = 23, 22 , 22, 21
Q) Which term of the sequence 23, 22 , 22, 21 … is the first negative term ?
12
12
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
To find: First negative term
Solution: a = first term = 23, 2nd term = 22 and
d = common difference = 22 - 23 = - 12
12
Then, an < 0
a + (n – 1)d < 0
23 + (n – 1) < 0
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
12
23 - + < 0n2
12
- < 047 2
n2
< 47 2
n2
47 < n
n > 47
Since, 48th is the natural number just greater than 47, therefore n = 48.
Thus, 48th term of the given sequence is the first negative term.
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Q) If the mth term of an AP be and nth term be , then Show that its (mn)th term is 1.
1n
1m
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Given: mth term =
nth term =
To prove: (mn)th term is 1
Solution: Let a = first term and d = common difference of the given AP.
am = a + (m – 1)d
= a + (m – 1)d am = , given … (1)1n
1n
and an = a + (n – 1)d
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
= a + (n – 1)d an = , given … (2) 1m
1m
Subtracting equation (2) from equation (1), we get
1n
1m
- = [a + (m – 1)d] - [a + (n – 1)d]
1n
- 1m = a + (m – 1)d - a – (n – 1)d
1n
- 1m
= (m – 1– n + 1)d
m - n mn
= (m– n)d
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
d = … (3) 1mn
Substituting the value of ‘d’ in equation (2), we get, 1 m = a + (n – 1) 1
mn
1m
- (n – 1) mn
= a
a = n – n + 1 mn
a = 1mn
….(4)
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
(mn)th term = amn = a + (mn – 1)d
amn = 1mn
[Using (3) and (4)]+ (mn – 1) 1mn
amn = 1mn
mn - 1 mn
+
amn = 1+ mn - 1 mn
mn mn
= = 1
Hence, the (mn)th term is 1
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
A negative term
from the given
sequence.
an = a + (n-1) d
Chapter : Arithmetic Progressions Website: www.letstute.com
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Problems based onArithmetic Progressions
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Part 4
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Chapter : Arithmetic Progressions Website: www.letstute.com
Q) The sum of three consecutive numbers in AP is - 6, and their product is + 64. Find the numbers.
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Given: S3 = - 6 Product of three consecutive numbers = 64
To Find: 3 consecutive numbers
Solution: Let the numbers be (a – d), a, (a + d)
Sum = - 6
(a – d) + a + (a + d) = - 6
a = - 2 …(1) 3a = - 6
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
(a – d) (a) (a + d) = + 64 a(a2 - d2) = + 64 -2[(-2)2 - d2] = + 64 [Using (1)] -2[4 - d2] = + 64 4 - d2 = - 32 d2 = 36
d = + 6
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Product = + 64
If d = – 6, the numbers are [– 2 – (– 6)], – 2 and [– 2 + (– 6)], i.e, 4, –2 and – 8
Hence, the required numbers are – 8, – 2, 4 or 4, – 2, – 8.
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
If d = + 6, the numbers are (–2 – 6), – 2 and (– 2 + 6), i.e, – 8, –2 and 4
Q) Find the four consecutive even number of terms in AP whose sum is 16 and the sum of whose squares is 84.
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Given: S4 = 16 Sum of squares of 4 consecutive even number of terms = 84
To Find: 4 consecutive even number of terms in AP
Solution: Let the four consecutive even number of terms in AP be a – 3d, a – d, a + d, a + 3d
Sum of the four consecutive even number of terms = 16
(a – 3d) + (a - d) + (a + d) + (a + 3d) = 16
a = 4 …(1)
4a = 16
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
a2 - 6ad + 9d2 + a2 - 2ad + d2 + a2 + 2ad + d2 + a2 +6ad + 9d2 = 84
(a – 3d)2 + (a - d)2 + (a + d)2 + (a + 3d)2 = 84
4a2 + 20d2 = 84
a2 + 5d2 = 21 42 + 5d2 = 21 [Using (1)]
5d2 = 5
d2 = 1
d = + 1
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Sum of the squares of the four consecutive even number ofterms in AP = 84
4(a2 + 5d2) = 84
If d = + 1 then the terms are (4 – 3), (4 – 1), (4 + 1), (4 + 3) i.e. 1, 3, 5, 7.
If d = - 1 then the terms are (4 + 3), (4 + 1), (4 - 1), (4 - 3) i.e. 7, 5, 3, 1.
Hence, the required consecutive even number of terms are 1, 3, 5, 7 or 7, 5, 3, 1.
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
The AP may be written as
a, (a + d), (a + 2d),…. (l - 2d), (l – d), l
Last term from the end is l = l – (1 – 1)d
Second term from the end is l – d = l – (2 – 1)dThird term from the end is l – 2d = l – (3 – 1)d
Fourth term from the end is l – 3d = l – (4 – 1)d … and so on.
The nth term from the end is l – (n – 1)d
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
nth term from the end
Let the AP be a, a + d, a + 2d, ….. l
First term = a, common difference = d and last term = l
Q)Find the 12th term from the end of the AP 3, 6, 9,…60
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Given: AP = 3, 6, 9,…60
To Find: 12th term from the end
Solution: d = common difference = 6 – 3 = 3 and the last term = l = 60
nth term from the end = l – (n – 1)d
= 60 – 11 x 3
Hence, the 12th term from the end of the AP 3, 6, 9… 60 is 27
12th term from the end = 60 – (12 – 1) x 3
= 60 – 33 = 27
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Finding out exact numbers when the sum and product of numbers is given.
nth term from the end using the formula l – (n – 1)d
Treasure
Problems based on Arithmetic Progressions
Chapter : Arithmetic Progressions Website: www.letstute.com
Chapter : Arithmetic Progressions Website: www.letstute.com
Now we know…
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Problems based onArithmetic Progressions