Summation
Arithmetic Progression
(AP)
An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference (d) of AP, it can be positive and negative
First term of an AP is (a)
1) General Form of an AP is
a, a + d, a + 2d, a + 3d,.....
2) nth Term of an AP
an = a + ( n - 1 ) d
Example
1) How many two- digit numbers are divisible by 3 ?
Solution :
The list of two-digit numbers divisible by 3 is :
12, 15, 18, .........99 .
an = a + ( n - 1 ) d
a = 12, d = 3, an = 99
99 = 12 + ( n - 1 ) 3
99 - 12 = ( n - 1 ) 3
87 = n - 1
3
29 = n - 1
29 + 1 = n
n = 30
So, there are 30, two-digit numbers divisible by 3.
2) Find the 10th term of the AP : 2, 7, 12,.......
Solution :
an = a + ( n - 1 ) d
Here a = 2, d = 7 - 2 = 5, and n = 10
a10 = 2 + ( 10 - 1 ) 5
= 2 + 9 × 5
= 2 + 45
= 47
Therefore, the 10th term of the given AP is 47 .
3) How many numbers between 1 to 50 are divisible by 3 ?
Solution :
Number divisible by 3 between 1 to 50 are
1×3 = 3
2×3 = 6
3×3 = 9
4×3 = 12........
16×3 = 48
AP is 3, 6, 9, 12,......... , 48
an = a + ( n - 1 ) d
here an = 48, a = 3, d = 6 - 3 = 3, n = ?
48 = 3 + ( n - 1 ) 3
48 - 3 = ( n - 1 ) 3
45/3 = n -1
15 + 1 = n
16 = n
Thus, 16 numbers between 1 to 50 to divisible by 3.
4) How many three digit numbers are divisible by 88 ?
Solution :
Three digit numbers divisible by 88 are
2×88 = 176
3×88 = 264
4×88 = 352
5×88 = 440.......
11×88 = 968.
AP is 176, 264, 352, .........., 968.
an = a + ( n - 1 ) d
an = 968, a = 176, d = 264 - 176 = 88,
968 = 176 + ( n - 1 ) 88
968 - 176 = ( n - 1 ) 88
792/88 = n - 1
9 + 1 = n
10 = n
Thus, Three digit numbers divisible by 88 are 10.
5) How many numbers are divisible by 5 and lying between 4000 and 5000 ?
Solution :
AP is 4000, 4005, 4010, .........., 5000.
an = a + ( n - 1 ) d
an = 5000, a = 4000, d = 4005 - 4000 = 5,
5000 = 4000 + ( n - 1 ) 5
5000 - 4000 = ( n - 1 ) 5
1000/5 = n - 1
200 + 1 = n
201 = n
Thus, 201 numbers divisible by 5 and lying between 4000 and 5000 .
6) The 6th term of an AP is 6 and the 16th term is 14. What is the 27th term ?
Solution :
an = a + ( n - 1 ) d
here
an = 6, n = 6,
6 = a + ( 6 - 1 ) d
6 = a + 5 d......................(1)
and
an = a + ( n - 1 ) d
here
an = 14, n = 16
14 = a + ( 16 - 1 ) d
14 = a + 15 d..................(2)
from eq. (2) - (1)
14 = a + 15 d
- 6 = a + 5 d
8 = 0 + 10 d
d = 8/10 = 4/5
put in (1)
6 = a + 5 × 4/5
6 = a + 4
a = 6 - 4
a = 2
again we use
an = a + ( n - 1 ) d
here
an = ?, n = 27, a = 2, d = 4/5
an = 2 + ( 27 - 1 ) 4/5
an = 2 + 26 × 4/5
an = (10 + 104)/5
an = 114/5
Thus, 27th term 114/5.
7) How many integers between 320 and 2000 are divisible by 3 ?
Solution :
Number divisible by 3 between 320 and 2000 are
AP is 321, 324, 327............., 1998.
an = a + ( n - 1 ) d
here an = 1998, a = 321, d = 324 - 327 = 3, n = ?
1998 = 321 + ( n - 1 ) 3
1998 - 321 = ( n - 1 ) 3
1677/3 = ( n -1)
559 + 1 = n
560 = n
Thus, 560 numbers between 321 to 2000 to divisible by 3.
8) How many integers between 30 and 320 are divisible by 3 ?
Solution :
Number divisible by 3 between 30 and 320 are
AP is 33, 36............., 318.
an = a + ( n - 1 ) d
here an = 318, a = 33, d = 36 - 33 = 3, n = ?
318 = 33 + ( n - 1 ) 3
318 - 33 = ( n - 1 ) 3
285/3 = ( n -1 )
95 + 1 = n
96 = n
Thus, 96 numbers between 30 to 320 to divisible by 3.
9) How many integers between 30 and 230 are divisible by 3 ?
Solution :
Number divisible by 3 between 30 and 230 are
AP is 33, 36.............,228 .
an = a + ( n - 1 ) d
here an = 228, a = 33, d = 36 - 33 = 3, n = ?
228 = 33 + ( n - 1 ) 3
228 - 33 = ( n - 1 ) 3
195/3 = ( n -1 )
65 + 1 = n
66 = n
Thus, 66 numbers between 30 to 230 to divisible by 3.
10) How many integers between 30 and 2020 are divisible by 3 ?
Solution :
Number divisible by 3 between 30 and 2020 are
AP is 33, 36.............,2019 .
an = a + ( n - 1 ) d
here an = 2019, a = 33, d = 36 - 33 = 3, n = ?
2019 = 33 + ( n - 1 ) 3
2019 - 33 = ( n - 1 ) 3
1986/3 = ( n -1 )
662 + 1 = n
663 = n
Thus, 663 numbers between 30 to 2020 to divisible by 3.
11) How many integers between 30 and 2030 are divisible by 3 ?
Solution :
Number divisible by 3 between 30 and 2030 are
AP is 33, 36.............,2028 .
an = a + ( n - 1 ) d
here an = 2028, a = 33, d = 36 - 33 = 3, n = ?
2028 = 33 + ( n - 1 ) 3
2028 - 33 = ( n - 1 ) 3
1995/3 = ( n -1 )
665 + 1 = n
666 = n
Thus, 666 numbers between 30 to 2030 to divisible by 3.
12) How many integers between 30 and 2040 are divisible by 3 ?
Solution :
Number divisible by 3 between 30 and 2040 are
AP is 33, 36.............,2037.
an = a + ( n - 1 ) d
here an = 2037, a = 33, d = 36 - 33 = 3, n = ?
2037 = 33 + ( n - 1 ) 3
2037 - 33 = ( n - 1 ) 3
2004/3 = ( n -1 )
668 + 1 = n
669 = n
Thus, 669 numbers between 30 to 2040 to divisible by 3.
13) How many integers between 40 and 2040 are divisible by 3 ?
Solution :
Number divisible by 3 between 40 and 2040 are
AP is 42, 45,.............,2037.
an = a + ( n - 1 ) d
here
2037 = 42 + (n - 1 ) 3
2037 - 42 = (n - 1) 3
1995/3 = ( n - 1 )
665 + 1 = n
n = 666
3) Sum of First n Terms of an AP
S = n [2a + ( n - 1 ) d ]
2
S = n [ a + a + ( n - 1 ) d ]
2
S = n [ a + an ]
2
Example:
1) Sum of first 20 natural numbers divisible 7 are
Solution:
Sn = n [ 2a + ( n - 1 ) d ]
2
where
a = 7,
n = 20,
d = 7,
Sn = ?
Sn = 20 [ 2 × 7 + ( 20 - 1 ) 7 ]
2
Sn = 10 [ 14 + 19 7 ]
Sn = 10 [ 14 + 98 ]
Sn = 10 [ 112 ]
Sn = 10 × 112
Sn = 1120
Thus, Sum of first 20 natural numbers divisible 7 are 1120 .
2)Find the sum of all natural numbers between 1 and 201 which divisible by 5
Solution:
AP 5, 10, 15, 20....... , 200
we use
An = a + ( n - 1 ) d
where
a = 5,
d = 10 - 5 = 5,
An = 200,
then
200 = 5 + ( n - 1 ) 5
200 - 5 = ( n - 1) 5
195 = n - 1
5
39 + 1 = n
n = 40
Sum of AP is
Sn = n [ 2a + ( n - 1 ) d ]
2
Sn = n [ a + { a + ( n - 1 ) d } ]
2
Sn = 40 [ 5 + 200 ]
2
Sn = 20 × 205
Sn = 4100
Thus, the sum of all natural numbers between 1 and 201 which divisible by 5 is 4100 .
2) Find the sum of the first 22 terms of the AP: 8, 3, -2,...
Solution :
Here,
a = 8 , d = 3 - 8 = -5, n = 22
Sn = n [ 2a + ( n - 1 ) d ]
2
Sn = 22 [ 2×8 + ( 22 - 1 ) ( - 5 ) ]
2
Sn = 11 [ 16 + 21 ( -5 ) ]
Sn = 11 [ 16 - 105 ]
Sn = 11 ( -89 )
Sn = - 979
So, the sum of the first 22 terms of AP is - 979 .
3) Sum of first 20 natural numbers divisible 7 are
Solution :
AP = 7, 14, 21,.........140.
Sn = n [ 2a + ( n - 1 ) d ]
2
where
n = 20, a = 7, d = 7
Sn = 20/2[2×7 + ( 20 - 1) 7 ]
Sn = 10 [ 14 + 19×7 ]
Sn = 10 [ 14 + 133 ]
Sn = 10 × 147
Sn = 1470
4) Find the sum of all natural numbers between 1 and 201 which divisible by 5
Solution:
AP = 5, 10, 15, .......200.
Sn = n [ 2a + ( n - 1 ) d ]
2
where a = 5, d = 5, n = 200/5 = 40
Sn = 40/2 [ 2×5 + ( 40 + 1 ) 5 ]
Sn = 20 [ 10 + 41×5 ]
Sn = 20 [ 10 + 205 ]
Sn = 20 × 215
Sn = 4300
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