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.

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