Arithmetic progressions - Problem based video part 4 for class 10th maths

Arithmetic Progressions
Problems based on
1 Chapter : Arithmetic Progressions Website: www.letstute.com

Q) Which term of the sequence 23, 22 , 22, 21 … is
the first negative term ?
Given : Sequence = 23, 22
ퟏ
ퟐ
, 22, 21
ퟏ
ퟐ
1
2
1
2
Problems based on
To find: First negative term
Chapter : Arithmetic Progressions Website: www.letstute.com

Problems based on
1
2
Solution: a = first term = 23, 2nd term = 22 and
d = common difference = 22 - 23 = - 1
2
1
2
Then, an < 0
 a + (n – 1)d < 0
 23 + (n – 1)
−ퟏ
ퟐ
< 0

n
2
Problems based on
1
2
 23 - + < 0
 47 - < 0
2
n
2
 47 <
2
n
2
 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
1
n
1
m
Q) If the mth term of an AP be and nth term be , then
Show that its (mn)th term is 1.
Given: mth term =
ퟏ
퐧
nth term =
ퟏ
퐦
To prove: (mn)th term is 1

Problems based on
Solution: Let a = first term and d = common difference of the
given AP.
am = a + (m – 1)d
 = a + (m – 1)d ∴ am 1 = , given … (1)
n
1
n
and an = a + (n – 1)d
1
m
1
m
 = a + (n – 1)d ∴ an = , given … (2)

Problems based on
Subtracting equation (2) from equation (1), we get
1
n
1
m
- = [a + (m – 1)d] - [a + (n – 1)d]

1
n
-
1
m = a + (m – 1)d - a – (n – 1)d
 1
n
- 1
m
= (m – 1– n + 1)d

m - n
mn
= (m– n)d
 d = 1 … (3)
mn

Substituting the value of ‘d’ in equation (2), we get,
1
m = a + (n – 1) 1
mn

1
m
- (n – 1)
mn
= a
 a = n – n + 1
mn
 a = 1
mn
….(4)
Problems based on

Problems based on
(mn)th term = amn = a + (mn – 1)d
 amn = 1
mn
1
mn
+ (mn – 1) [Using (3) and (4)]
 amn = 1
mn
mn - 1
mn
+
 amn = 1+ mn - 1
mn
mn
mn
= = 1
Hence, the (mn)th term is 1

Now we know…
Problems based on
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Some more problems based on
Part 4
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Problems based on
Q) The sum of three consecutive numbers in AP is - 6,
and their product is + 64. Find the numbers.
Given: S3 = - 6
Product of three consecutive numbers = 64
To Find: 3 consecutive numbers

Problems based on
Solution: Let the numbers be (a – d), a, (a + d)
Sum = - 6
 (a – d) + a + (a + d) = - 6
 3a = - 6
 a = - 2 …(1)

Problems based on
Product = + 64
 (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
If d = + 6, the numbers are (–2 – 6), – 2 and (– 2 + 6),
i.e, – 8, –2 and 4
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
Q) Find the four consecutive even number of terms in AP
whose sum is 16 and the sum of whose squares is 84.
Given: S4 = 16
Sum of squares of 4 consecutive even number of terms = 84
To Find: 4 consecutive even number of terms in AP

Problems based on
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
 4a = 16
 a = 4 …(1)

Problems based on
Sum of the squares of the four consecutive even number of
terms in AP = 84
 (a – 3d)2 + (a - d)2 + (a + d)2 + (a + 3d)2 = 84
a2 - 6ad + 9d2 + a2 - 2ad + d2 + a2 + 2ad + d2 + a2 +6ad + 9d2 = 84
 4a2 + 20d2 = 84
4(a2 + 5d2) = 84

 a2 + 5d2 = 21
 42 + 5d2 = 21 [Using (1)]
 5d2 = 5
 d2 = 1
 d = + 1

Problems based on
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
 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
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)d
Third 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
Q)Find the 12th term from the end of the AP 3, 6, 9,…60
Given: AP = 3, 6, 9,…60
To Find: 12th term from the end

Problems based on
Solution: d = common difference = 6 – 3 = 3 and
the last term = l = 60
nth term from the end = l – (n – 1)d
∴ 12th term from the end = 60 – (12 – 1) x 3
= 60 – 11 x 3
= 60 – 33 = 27
Hence, the 12th term from the end of the AP 3, 6, 9… 60 is 27

Problems based on
Treasure
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

Now we know…
Problems based on
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Arithmetic progressions - Problem based video part 4 for class 10th maths

Arithmetic progressions - Problem based video part 4 for class 10th maths

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Arithmetic progressions - Problem based video part 4 for class 10th maths

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