Arithmatic progression for Class 10 by G R Ahmed
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An arithmetic progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. The constant difference is called the common difference. The sum of the first n terms of an AP can be calculated using the formula: Sum = n/2 * (first term + last term). Several word problems are presented involving finding sums, terms, or properties of AP sequences.
![Q.Find the sum of all 3 digit numbers which
leave remainder 3 when divided by 5.
Ans: 103, 108……….998
here, a + (n-1)d = 998
⇒ 103 + (n-1)5 = 998
⇒ n = 180
S180 = 180/2[103 + 998]
= 90 x 1101
S180 = 99090](https://image.slidesharecdn.com/arithmaticprogression-200429045035/85/Arithmatic-progression-for-Class-10-by-G-R-Ahmed-10-320.jpg)
![Q.Find the sum of all natural numbers amongst first one
thousand numbers which are neither divisible 2 or by 5
Ans: Sum of all natural numbers in first 1000 integers which
are not divisible by 2 i.e.
sum of odd integers.
1 + 3 + 5 + ………. + 999
Here n = 500
S500 = 500/2 [1 + 999]
= 2,50,000
No’s which are divisible by 5
5 + 15 + 25 …….. + 995 here n = 100
S100 =100/2 [5 + 995]
= 50 x 1000 = 50000
∴ Required sum = 250000 – 50,000
= 200000](https://image.slidesharecdn.com/arithmaticprogression-200429045035/85/Arithmatic-progression-for-Class-10-by-G-R-Ahmed-12-320.jpg)
![A thief runs away from a police station with a uniform speed of 100m/minutes. After one minute a
policeman runs behind the thief to catch him at speed of 100m/minute in first minute & increases his
speed 10m each succeeding minute. After how many minutes, the policeman will catch the thief?
Let the police catch the thief in ‘n’ minutes
Since the thief ran 1 min before the police start running.
Time taken by the thief before he was caught = (n + 1) min
Distance travelled by the thief in (n+1) min = 100(n+1)m
Given speed of policeman increased by 10m per minute.
Speed of police in the 1st min = 100m/min
Speed of police in the 2nd min = 110m/min
Speed of police in the 3rd min = 120m/min
Hence 100, 110, 120… are in AP
Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d]
= (n/2)[2 x 100 +(n – 1)10]
After the thief was caught by the police,
Distance traveled by the thief = distance travelled by the police
100(n+1)= (n/2)[2 x 100 +(n-1)10]
200(n + 1) = n[200 + 10n – 10]
200n + 200= 200n + n(10n – 10 )
200 = n(n – 1)10](https://image.slidesharecdn.com/arithmaticprogression-200429045035/85/Arithmatic-progression-for-Class-10-by-G-R-Ahmed-18-320.jpg)
Arithmatic progression for Class 10 by G R Ahmed
- 1. Arithmetic Progression (A.P): A sequence of numbers is said to be in A.P when difference between two consecutive terms is constant . This constant is called common difference. Ex. Let us check the sequence of the numbers 2, 4, 6, 8, 10………………………… here 1st term(a)=2 and common difference(d)= 2 If a the first term and d the common difference of the A.P then,
- 4. The sum of A.P
- 5. nn aa n S 1 2 first term last term Let’s find the sum of 1 + 3 +5 + . . . + 59 12 nThe common difference is 2 and the first term is one so: Set this equal to 59 to find n. Remember n is the term number. 2n - 1 = 59 n = 30 So there are 30 terms to sum up.
- 6. Selection of Terms in an Arithmetic Progression (i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a - d, a and a + d. Here common difference is d. (ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a - 3d, a - d, a + d and a + 3d. Here the common difference is 2d
- 7. (iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a - 2d, a - d, a, a + d and a + 2d. Here common difference is d. (iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a - 5d, a - 3d, a - d, a + d, a + 3d and a + 5d. Here common difference is 2d.
- 9. Q. The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term. Ans: 4th term= 0 ⇒ a + 3d = 0 25th term=a +24d=3(a+10d)-2a-6d = 3x11th term-2(a+3d) = 3x11th term-2x 0 = 3 x 11th term
- 10. Q.Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5. Ans: 103, 108……….998 here, a + (n-1)d = 998 ⇒ 103 + (n-1)5 = 998 ⇒ n = 180 S180 = 180/2[103 + 998] = 90 x 1101 S180 = 99090
- 11. Q.Find the value of x if 2x + 1, x² + x +1, 3x² - 3x +3 are consecutive terms of an AP. Ans: a2 –a1 = a3 –a2 ⇒ x² + x + 1-2x - 1 = 3x² – 3x + 3- X2 -x-1 x² - x = 2x² – 4x + 2 ⇒ x² - 3x + 2 = 0 ⇒ (x -1) (x – 2) = 0 ⇒ x = 1 or x = 2
- 12. Q.Find the sum of all natural numbers amongst first one thousand numbers which are neither divisible 2 or by 5 Ans: Sum of all natural numbers in first 1000 integers which are not divisible by 2 i.e. sum of odd integers. 1 + 3 + 5 + ………. + 999 Here n = 500 S500 = 500/2 [1 + 999] = 2,50,000 No’s which are divisible by 5 5 + 15 + 25 …….. + 995 here n = 100 S100 =100/2 [5 + 995] = 50 x 1000 = 50000 ∴ Required sum = 250000 – 50,000 = 200000
- 13. Q. The angles of a quadrilateral are in A.P. whose common difference is 15°. Find the angles.
- 14. Q. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
- 18. A thief runs away from a police station with a uniform speed of 100m/minutes. After one minute a policeman runs behind the thief to catch him at speed of 100m/minute in first minute & increases his speed 10m each succeeding minute. After how many minutes, the policeman will catch the thief? Let the police catch the thief in ‘n’ minutes Since the thief ran 1 min before the police start running. Time taken by the thief before he was caught = (n + 1) min Distance travelled by the thief in (n+1) min = 100(n+1)m Given speed of policeman increased by 10m per minute. Speed of police in the 1st min = 100m/min Speed of police in the 2nd min = 110m/min Speed of police in the 3rd min = 120m/min Hence 100, 110, 120… are in AP Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d] = (n/2)[2 x 100 +(n – 1)10] After the thief was caught by the police, Distance traveled by the thief = distance travelled by the police 100(n+1)= (n/2)[2 x 100 +(n-1)10] 200(n + 1) = n[200 + 10n – 10] 200n + 200= 200n + n(10n – 10 ) 200 = n(n – 1)10
- 19. n(n – 1) – 20 = 0 n2 – n– 20 = 0 n2 – 5n + 4n – 20 = 0 n(n – 5) + 4(n – 5) = 0 (n – 5) (n+4) = 0 (n – 5) = 0 or (n + 4) = 0 n= 5 or n= -4 Hence n= 5 since n cannot be negative Therefore the time taken by the policeman to catch the thief = 5minutes