Arithmetic Progression (AP) |
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Types of arithmetic progressions |
What is a progression? |
| In a sequence of numbers, if every term except the first one has a common relation to the preceding number, then that sequence is called a progression. For example, the sequence 1, 3, 5, 7, . . . n is called a progression because the series has a common relation, i.e., if you add 2 to any term, you get the succeeding number. We denote the terms in the sequence as t1 t2 t3, . . . tn for the first term, second term, third term and so on, and the last term is tn. |
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There are three types of progressions |
1. Arithmetic progression
2. Geometric progression
3. Harmonic progression |
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| Let’s study them in detail. |
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Arithmetic progression |
| Consider the following examples : |
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- 1, 3, 5, 7, 9, . . . .
- 2, 4, 6, 8, 10, . . .
- 4, 7, 10, . . .
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In each of the above examples, we find an arrangement of numbers in a definite order. Also, the terms follow a certain rule. In the first and second examples, by subtracting 2 we get the preceding number. In the third one, we need to subtract 3.
Therefore, an arithmetic progression (A.P.) is a sequence in which each term, except the first, is obtained by adding a fixed number to the term immediately preceding it. The first term is noted as “a”. This fixed number is called the common difference and is usually denoted by “d”. |
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For example, in our example series 1, 3, 5, 7, 9, . . .
a = 1, which is the first term
d is 2, which is the difference of two consecutive numbers
(3–1)(5–3)(7–5)(9–7) and so on.
Therefore, in a sequence t1 t2 t3, . . . tn
the common difference, d = t2 – t1 = t3 – t2. . . tn – tn – 1 |
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Examples |
16, 19, 22, 25, . . . here, a = 16; d = 3 = 19 - 16
21, 16, 11, 6, 1, . . . here, a = 21; d = – 5 = 16 - 21
– 1, – 4, – 7, – 10, . . . here, a = – 1; d = – 3 = -4-(-1)
x–3b, x+b, x+5b, x+9b, here, a = x – 3b; d = 4b=x+b-(x+3b)
a – 3b, 2a – 5b, 3a – 7b, here, a = a – 3b; d = a – 2b=(2a-5b)-(a-3b) |
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The nth term or general term of an arithmetic progression |
Let us consider an A.P. with first term “a” and common difference “d”.
The second term is obtained by adding “d” to the first term “a”.
Thus, t2 = a + d; similarly, t3 is obtained by adding “d” to t2.
Thus, t3 = t2 + d = a + d + d = a + 2d.
We write these and a few more terms as shown below:
First term t1= a = a + (1 – 1)d
Second term t2 = a + d = a + (2 – 1)d
Third term t3 = a + 2d = a + (3 – 1)d
Fourth term t4 = a + 3d = a + (4 – 1)d
Do you see a pattern?
We see that in any term, the coefficient of d is always one less than the number of terms in the series.
t16 = a + (16 – 1)d = a + 15d
In general, the nth term is tn = a + (n – 1)d |
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Example |
1) Find the 18th and 23rd term and the general term of the A.P. 16, 11, 6, 1, -4, -9.
Solution:
Here, a = 16; d = -5
Thus, t18 = a + (18 - 1)d
= 16 + 17(-5)
= -69
t23 = a + (23 - 1)d
= 16 + 22(-5)
= -94
And tn= a + (n - 1)d
= 16 + (n - 1)(-5)
= 16 - 5n + 5
= 21 - 5n
2) The first term of an A.P. is -2 and the 10th term is 16. Determine the 15th term.
Solution:
The formula for the general term of an A.P. involves two variables, “a” and “d”. In this problem, we are given two conditions, namely
t1 = -2 and t10 = 16.
Thus, t1 = a = -2; t10 = a + 9d = 16
By substituting the value of “a” in t10, we get:
t10 = -2 + 9d = 16 or 9d = 18 ∴ d = 2
So, t15 = a + 14d = -2 + 14(2) = -2 + 28 = 26. |
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The sum of the first n terms of an arithmetic progression |
Carl Friedrich Gauss, the great German mathematician, was in elementary school. One day, his teacher gave the class a problem of finding the sum of the first 100 natural numbers.
While the rest of the class was struggling with the problem, Gauss gave the answer in no time.
1) He wrote the first 100 natural numbers as given below. S denotes the sum to be determined.
S = 1 + 2 + 3 +. . . + 97 + 98 + 99 + 100
Reversing S = 100 + 99 + 98 +, . . . + 4 + 3 + 2 + 1
–––––––––––––––––––––––––––––––––––––––––––
Adding 2S = 101 + 101 + 101 +, . . . + 101 + 101 + 101 + 101
2S = 101 * 100
101 * 100
∴ S = ––––––––
2
= 101 * 50
= 5050.
We now provide a method of finding the sum of n terms of an A.P.
Let the first term of an A.P. be “a” and the common difference “d”.
Let Sn denote the sum of the first “n” natural numbers. |
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3) Find the sum of the first 100 natural numbers.
Solution:
Here the first term a = 1; common difference d = 1 and the number of terms n = 100
100 [2(1) + (100–1) *1]
Thus, S100 = –––––––––––––––––––––
2
= 50(2 + 99)
= 50 *101
= 5050
4) Find the sum of all the natural numbers between 1 and 100 that are multiples of 3.
Solution:
Here a = 3; d = 3; and l = 99, where l = a + (n-1)d
We must find n.
We have l = a + (n – 1)d
i.e., 99 = 3 + (n – 1) 3
Or 99 = 3 + 3n – 3 = 3n
 n = 99/3 = 33
33(3 + 99) (33 *102)
And S33 = –––––––– = –––––––– = 1683
2 2 |
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I) Find the following |
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1. |
Find the 25th term of an A.P. whose 8thterm is 17 and the 19thterm is 39 |
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2. |
Find d and write the next two terms of the following A.P
21, 25, 29, . . . |
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3. |
10, 17, 24, . . . |
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4. |
0, – 3, – 6, – 9, . . . |
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5. |
Find t10 when a = 3; d = 2 |
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6. |
Find t18 when a = 16; d = – 5 |
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7. |
Find t5 when a = –1; d = – 3 |
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8. |
Find t11 when a = 1; d = 2 |
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9. |
The sum of three terms of an A.P. is 36 while their product is 1620. Find the A.P. |
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10. |
Determine the sum of first 35 terms of an A.P. if t2 = 2 and t7 = 22. |
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II) Choose the right answer |
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11. |
In the arithmetic series 10, 17, 24, . . . the value of d is:
i) 7 ii) –7 iii) 33 |
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12. |
The value of d in the A.P. 0, – 3, – 6, – 9, . . . is:
i) 3 ii) –3 iii) –1 |
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13. |
The first five terms in the series whose nth term is tn = n2 + 1 are:
i) 0, 2, 5, 10, 17 ii) 1, 2, 5, 10, 17 iii) 2, 5, 10, 17, 26 |
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14. |
The second term in the series whose nth term is tn = n(n + 1) is:
i) 2 ii) 6 iii) 12 |
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15. |
n(n + 3)
The 17thterm in the series if tn = ————- is:
(n + 2)
i) 33/19 ii) 340/19 iii) 340/20 |
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16. |
The 10thterm of the A.P. 13, 8, 3, – 2, . . . is
(a) – 32 (b) 32 (c) – 57 (d) – 52 |
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17. |
If a, b, c are in A.P. then b =
(a) (a + c)/2 (b) (a – c)/2 (c) ac/2 (d) 2/ac |
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18. |
If a, b, c are in A.P. then b – a =
(a) 1 (b) a – c (c) c – b (d) a + b |
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19. |
Sum of the terms 1, 2, 3, 4, . . . ,n is
(a) n(n – 1)/2 (b) n(n + 1)/2 (c) n2(n + 1)2/4 (d) n2(n + 1)/2 |
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20. |
If the nthterm of an A .P is 2n + 5 ,its first term is
(a) 7 (b) ½ (c) – 2 (d) 2 |
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Answers to Practice Problems |
I) Find the following |
| 1. |
Find the 25th term of an A.P. whose 8th term is 17 and the 19th term is 39
Solution:
Let the first term be “a” and common difference d.
Then t8 = a + (8 – 1)d = a + 7d
t19= a + (19 – 1)d = a + 18d
It is given that t8= 17 and t19 = 39
We get:
a + 7d = 17 ———— (1)
a + 18d = 39 ———–(2)
By subtracting (1) from (2) we get 11d = 22 d = 2
By substituting this in either (1) or (2) we get a = 3
Now t25 = a + (25 – 1)d = a + 24d = 3 + (24 2) = 3 + 48 = 51 |
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2. |
Find d and write the next two terms of the following A.P
21, 25, 29, . . .
Here d = 4 and the next two numbers are: 33, 37 |
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3. |
10, 17, 24, . . .
Here d = 7 and the next numbers are: 31, 38 |
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4. |
0, – 3, – 6, – 9, . . .
Here d = – 3 and the next numbers: – 12, – 15 |
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5. |
Find t10 when a = 3; d = 2
Solution:
tn = a + (n – 1)d
t10 = 3 + (10 – 1)2
= 3 + (9 2)
= 3 + 18 = 22. |
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6. |
Find t18 when a = 16; d = – 5
Solution:
t18 = 16 + (18 – 1)(– 5)
= 16 + (17 – 5)
= 16 + ( – 85)
= – 69. |
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7. |
Find t5 when a = – 1; d = – 3
Solution:
t5 = – 1 + ( 5 – 1)(– 3)
= – 1 + 4(– 3)
= – 1 – 12 = – 13. |
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8. |
Find t11 when a = 1; d = 2
Solution:
t11 = 1 + (11 – 1)2
= 1 + 10 * 2
= 1 + 20 = 21. |
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