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Arithmetic Progressions 
Problems based on 
Arithmetic Progressions 
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 
Arithmetic Progressions 
To find: First negative term 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
n 
2 
Problems based on 
Arithmetic Progressions 
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. 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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) 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
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 
Arithmetic Progressions 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
(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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Arithmetic progressions - Problem based video part 4 for class 10th maths
Now we know… 
Problems based on 
Arithmetic Progressions 
Please visit www.letstute.com to take a test 
Chapter : Arithmetic Progressions Website: www.letstute.com
Next video…. 
Some more problems based on 
Arithmetic Progressions 
Part 4 
Please visit www.letstute.com to view the next video 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
Solution: Let the numbers be (a – d), a, (a + d) 
Sum = - 6 
 (a – d) + a + (a + d) = - 6 
 3a = - 6 
 a = - 2 …(1) 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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. 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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) 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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. 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
 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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Problems based on 
Arithmetic Progressions 
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 
Chapter : Arithmetic Progressions Website: www.letstute.com
Now we know… 
Problems based on 
Arithmetic Progressions 
Next slide - Problem based Video 
Please visit www.letstute.com to take a test 
Chapter : Arithmetic Progressions Website: www.letstute.com
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

  • 1. Arithmetic Progressions Problems based on Arithmetic Progressions 1 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 2. 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 Arithmetic Progressions To find: First negative term Chapter : Arithmetic Progressions Website: www.letstute.com
  • 3. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 4. n 2 Problems based on Arithmetic Progressions 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. Chapter : Arithmetic Progressions Website: www.letstute.com
  • 5. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 6. Problems based on Arithmetic Progressions 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) Chapter : Arithmetic Progressions Website: www.letstute.com
  • 7. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 8. 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 Arithmetic Progressions Chapter : Arithmetic Progressions Website: www.letstute.com
  • 9. Problems based on Arithmetic Progressions (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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 11. Now we know… Problems based on Arithmetic Progressions Please visit www.letstute.com to take a test Chapter : Arithmetic Progressions Website: www.letstute.com
  • 12. Next video…. Some more problems based on Arithmetic Progressions Part 4 Please visit www.letstute.com to view the next video Chapter : Arithmetic Progressions Website: www.letstute.com
  • 13. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 14. Problems based on Arithmetic Progressions Solution: Let the numbers be (a – d), a, (a + d) Sum = - 6  (a – d) + a + (a + d) = - 6  3a = - 6  a = - 2 …(1) Chapter : Arithmetic Progressions Website: www.letstute.com
  • 15. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 16. Problems based on Arithmetic Progressions 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. Chapter : Arithmetic Progressions Website: www.letstute.com
  • 17. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 18. Problems based on Arithmetic Progressions 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) Chapter : Arithmetic Progressions Website: www.letstute.com
  • 19. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 20. Problems based on Arithmetic Progressions 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. Chapter : Arithmetic Progressions Website: www.letstute.com
  • 21. Problems based on Arithmetic Progressions  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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 22. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 23. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 24. Problems based on Arithmetic Progressions 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 Chapter : Arithmetic Progressions Website: www.letstute.com
  • 25. Now we know… Problems based on Arithmetic Progressions Next slide - Problem based Video Please visit www.letstute.com to take a test Chapter : Arithmetic Progressions Website: www.letstute.com

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