**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 t_{1} t_{2} t_{3}, . . . 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**

- Arithmetic progression
- progression
- Harmonic progression

Let’s study them in detail.

- 1, 3, 5, 7, 9, . . .
- 2, 4, 6, 8, 10, . . .
- 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 t_{1} t_{2} t_{3}, . . . tn

the common difference, d = t_{2} – t_{1} = t_{3} – t_{2}. . . tn – tn – 1

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)

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, t_{2} = a + d; similarly, t_{3} is obtained by adding “d” to t_{2}.

Thus, t_{3} = t_{2} + d = a + d + d = a + 2d.

We write these and a few more terms as shown below:

First term t_{1}= a = a + (1 – 1)d

Second term t_{2} = a + d = a + (2 – 1)d

Third term t_{3} = a + 2d = a + (3 – 1)d

Fourth term t_{4} = 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.

t_{16} = a + (16 – 1)d = a + 15d

In general, the n_{th} term is tn = a + (n – 1)d

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, t_{18} = a + (18 - 1)d

= 16 + 17(-5)

= -69

t_{23} = 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:**

t_{1} = -2 and t_{10} = 16.

Thus, t_{1} = a = -2; t_{10} = a + 9d = 16

By substituting the value of “a” in t10, we get:

t_{10} = -2 + 9d = 16 or 9d = 18 ∴ d = 2

So, t_{15} = a + 14d = -2 + 14(2) = -2 + 28 = 26.

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

- Find the 25
^{th}term of an A.P. whose 8^{th}term is 17 and the 19^{th}term is 39 - Find d and write the next two terms of the following A.P 21, 25, 29, . . .
- 10, 17, 24, . . .
- 0, – 3, – 6, – 9, . . .
- Find t
_{10}when a = 3; d = 2 - Find t
_{18}when a = 16; d = – 5 - Find t
_{5}when a = –1; d = – 3 - Find t
_{11}when a = 1; d = 2 - The sum of three terms of an A.P. is 36 while their product is 1620. Find the A.P.
- Determine the sum of first 35 terms of an A.P. if t
_{2}= 2 and t_{7}= 22. - In the arithmetic series 10, 17, 24, . . . the value of d is:
- The value of d in the A.P. 0, – 3, – 6, – 9, . . . is:
- The first five terms in the series whose nth term is tn = n2 + 1 are:
- The second term in the series whose n
^{th}term is t_{n}= n(n + 1) is: -
n(n + 3)

The 17^{th}term in the series if tn = ————- is:

(n + 2)

i) 33/19 ii) 340/19 iii) 340/20 - The 10
^{th}term of the A.P. 13, 8, 3, – 2, . . . is - If a, b, c are in A.P. then b =
- If a, b, c are in A.P. then b – a =
- Sum of the terms 1, 2, 3, 4, . . . ,n is
- If the n
^{th}term of an A .P is 2n + 5 ,its first term is

i) 7 ii) –7 iii) 33

i) 3 ii) –3 iii) –1

i) 0, 2, 5, 10, 17 ii) 1, 2, 5, 10, 17 iii) 2, 5, 10, 17, 26

i) 2 ii) 6 iii) 12

(a) – 32 (b) 32 (c) – 57 (d) – 52

(a) (a + c)/2 (b) (a – c)/2 (c) ac/2 (d) 2/ac

(a) 1 (b) a – c (c) c – b (d) a + b

(a) n(n – 1)/2 (b) n(n + 1)/2 (c) n^{2}(n + 1)^{2}/4 (d) n^{2}(n + 1)/2

(a) 7 (b) ½ (c) – 2 (d) 2

**I) Find the following**

- 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 t_{8}= a + (8 – 1)d = a + 7d

t_{19}= a + (19 – 1)d = a + 18d

It is given that t_{8}= 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 t_{25}= a + (25 – 1)d = a + 24d = 3 + (24 2) = 3 + 48 = 51

- 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 - 10, 17, 24, . . .

Here d = 7 and the next numbers are: 31, 38 - 0, – 3, – 6, – 9, . . .

Here d = – 3 and the next numbers: – 12, – 15 - Find t
_{10}when a = 3; d = 2

**Solution:**

t_{n}= a + (n – 1)d

t_{10}= 3 + (10 – 1)2

= 3 + (9 2)

= 3 + 18 = 22.

- Find t
_{18}when a = 16; d = – 5

**Solution:**

t_{18}= 16 + (18 – 1)(– 5)

= 16 + (17 – 5)

= 16 + ( – 85)

= – 69. - Find t
_{5}when a = – 1; d = – 3

**Solution:**

t_{5}= – 1 + ( 5 – 1)(– 3)

= – 1 + 4(– 3)

= – 1 – 12 = – 13. - Find t
_{11}when a = 1; d = 2

**Solution:**

t_{11}= 1 + (11 – 1)2

= 1 + 10 * 2

= 1 + 20 = 21.

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