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, . . . t_n for the first term, second term, third term and so on, and the last term is t_n.

There are three types of progressions

1. Arithmetic progression
2. progression
3. Harmonic progression

Let’s study them in detail.

Arithmetic progression

1. 1, 3, 5, 7, 9, . . .
2. 2, 4, 6, 8, 10, . . .
3. 4, 7, 10, . . .

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”.

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

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)

The n_th 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 n_th term is tn = a + (n – 1)d

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 t_n= 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 10^th term is 16. Determine the 15^th term.

Solution:

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.

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 S_n denote the sum of the first “n” natural numbers.

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

I)   Find the following

1. Find the 25^th term of an A.P. whose 8^th term is 17 and the 19^th term is 39
2. Find d and write the next two terms of the following A.P 21, 25, 29, . . .
3. 10, 17, 24, . . .
4. 0, – 3, – 6, – 9, . . .
5. Find t₁₀ when a = 3; d = 2
6. Find t₁₈ when a = 16; d = – 5
7. Find t₅ when a = –1; d = – 3
8. Find t₁₁ when a = 1; d = 2
9. The sum of three terms of an A.P. is 36 while their product is 1620. Find the A.P.
10. Determine the sum of first 35 terms of an A.P. if t₂ = 2 and t₇ = 22.

II)   Choose the right answer

11. In the arithmetic series 10, 17, 24, . . . the value of d is:

i)   7        ii)   –7        iii)   33



12. The value of d in the A.P. 0, – 3, – 6, – 9, . . . is:

i)   3        ii)   –3        iii)   –1



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



14. The second term in the series whose n^th term is t_n = n(n + 1) is:

i)   2        ii)   6        iii) 12



15.                                                   n(n + 3)
    The 17^thterm in the series if tn =  ————- is:
                                                       (n + 2)
    i) 33/19          ii) 340/19          iii) 340/20


16. The 10^th term of the A.P. 13, 8, 3, – 2, . . . is

(a)   – 32        (b)   32        (c)   – 57        (d)   – 52



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



18. If a, b, c are in A.P. then b – a =

(a)  1        (b)  a – c        (c)  c – b        (d)  a + b



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



20. If the n^th term of an A .P is 2n + 5 ,its first term is

(a)   7        (b)   ½        (c)   – 2        (d)   2

Answers to Practice Problems

I)   Find the following

1. Find the 25^th term of an A.P. whose 8^th term is 17 and the 19^th 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
   
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
3. 10, 17, 24, . . .
   Here d = 7 and the next numbers are: 31, 38
4. 0, – 3, – 6, – 9, . . .
   Here d = – 3 and the next numbers: – 12, – 15
5. Find t₁₀ when a = 3; d = 2
   Solution:
   t_n = a + (n – 1)d
   t₁₀ = 3 + (10 – 1)2
   = 3 + (9 2)
   = 3 + 18 = 22.
   
6. Find t₁₈ when a = 16; d = – 5
   Solution:
   t₁₈ = 16 + (18 – 1)(– 5)
   = 16 + (17 – 5)
   = 16 + ( – 85)
   = – 69.
7. Find t₅ when a = – 1; d = – 3
   Solution:
   t₅ = – 1 + ( 5 – 1)(– 3)
   = – 1 + 4(– 3)
   = – 1 – 12 = – 13.
8. Find t₁₁ when a = 1; d = 2
   Solution:
   t₁₁ = 1 + (11 – 1)2
   = 1 + 10 * 2
   = 1 + 20 = 21.