Rahul applied for a job and got selected. He is offered a starting monthly salary of $20000, with an annual increment of $500. His salary (in $) for the 1st, 2nd, 3rd,â€¦ years will be 20000, 20500, 21000,â€¦..

In a school, five cash rewards are given to the students for their academic performance. If the first prize is $500 and the subsequent rewards are $50 less than the preceding reward, the value of each reward (in $) will be 500, 450, 400, 350, 300.

In a fixed deposit scheme, the amount becomes \frac{5}{4}

4

5

â€‹

times itself after every 2 years, then the maturity amount of an investment of $8000 after 2, 4, 6 and 8 years (in $) will be: 10000, 12500, 15625, 19531.25 (in $) .

In the above examples we have seen some pattern being followed. We can see that succeeding terms are obtained by adding with a fixed number, subtracting with a fixed number, multiplying with a fixed number, and so on.

Now we will see how to find nth terms of such arithmetic series.

Common difference of arithmetic sequence:

The fixed number that is added to preceding term to obtain the next term is called the common difference of arithmetic sequence. It can be positive, negative and even zero.

If we denote first term of an AP by {a_1}a

1

â€‹

, second term by {a_2}a

2

â€‹

and so on and nth term by {a_n}a

n

â€‹

and the common difference by d, then the AP becomes {a_1},\,{a_2},\,{a_3}, \ldots ,{a_n}a

1

â€‹

,a

2

â€‹

,a

3

â€‹

,â€¦,a

n

â€‹

.

Therefore, {a_2} – {a_1} = {a_3} – {a_2} = \cdots = {a_n} – {a_{n – 1}} = da

2

â€‹

âˆ’a

1

â€‹

=a

3

â€‹

âˆ’a

2

â€‹

=â‹¯=a

n

â€‹

âˆ’a

nâˆ’1

â€‹

=d.

The general form of arithmetic sequence can be written as a,\,a + d,\,a + 2d,a + 3d,…..a,a+d,a+2d,a+3d,….. where a represents first term of arithmetic sequence and d denotes the common difference.

If we know the first term and the common difference, we can find the arithmetic sequence, because we know that in arithmetic sequence, every succeeding term is obtained by adding d to the preceding term. We can also find d of arithmetic sequence by subtracting any term from its succeeding term.

nth term of an arithmetic sequence:

Let us consider the situation where Rahul applied for a job and got selected. He was offered a starting monthly salary of $20000, with an annual increment of $500. Let us check what will be his monthly salary in the fifth year:

Let us first see his monthly salary for the second year,

{\rm{Salary\ for\ second\ year}} = \left( {20000{\rm{ }} + {\rm{ }}500} \right)Salary for second year=(20000+500)

{\rm{Salary\ for\ third\ year}} = \;\left( {20000 + 500 + 500} \right)Salary for third year=(20000+500+500)

= \left( {20000 + 2 \times 500} \right)=(20000+2Ã—500)

= \left[ {20000 + \left( {3 – 1} \right) \times 500} \right]=[20000+(3âˆ’1)Ã—500]

= 21000=21000

{\rm{Salary\ for\ fourth\ year}} = \left( {21000 + 500} \right)Salary for fourth year=(21000+500)

= \left( {20000 + 500 + 500 + 500} \right)=(20000+500+500+500)

= \left( {20000 + 3 \times 500} \right)=(20000+3Ã—500)

= \left[ {20000 + (4 – 1) \times 500} \right]=[20000+(4âˆ’1)Ã—500]

= 21500=21500

{\rm{Salary\ for\ fifth\ year}} = \left( {21500 + 500} \right)Salary for fifth year=(21500+500)

= \left( {20000 + 500 + 500 + 500 + 500} \right)=(20000+500+500+500+500)

= \left( {20000 + 4 \times 500} \right)=(20000+4Ã—500)

= \left[ {20000 + (5 – 1) \times 500} \right]=[20000+(5âˆ’1)Ã—500]

= 22000=22000

Observe we are getting a list of numbers,

20000, 20500,21000, 21500, 22000,â€¦â€¦

Now looking at this pattern, we can find his salary for any year. Let us try to find for the 15th year:

{\rm{Salary\ for\ 1}}{{\rm{5}}^{th}}{\rm{ year}} = {\rm{Salary}}\,{\rm{for}}\,{\rm{1}}{{\rm{4}}^{{\rm{th}}}}\,{\rm{year}}\,{\rm{ + }}\,{\rm{500}}Salary for 15

th

year=Salaryfor14

th

year+500

{\rm{ = }}\left( {20000 + 13 \times 500} \right)\, + \,500=(20000+13Ã—500)+500

= \left( {20000 + 14 \times 500} \right)=(20000+14Ã—500)

= \left[ {20000 + (15 – 1) \times 500} \right]=[20000+(15âˆ’1)Ã—500]

= 27000=27000

Which is First salary + (15-1) X Annual increment.

This example gives us some idea to write nth term.

Let {a_1},\,{a_2},\,{a_3},…….,{a_n}a

1

â€‹

,a

2

â€‹

,a

3

â€‹

,…….,a

n

â€‹

be an arithmetic sequence whose first term {a_1}a

1

â€‹

is a and the common difference is d.

Then,

Second term is {a_2} = a + d = a + \left( {2 – 1} \right)da

2

â€‹

=a+d=a+(2âˆ’1)d

Third term is {a_3} = {a_2} + d = \left( {a + d} \right) + d = a + 2d = a + \left( {3 – 1} \right)da

3

â€‹

=a

2

â€‹

+d=(a+d)+d=a+2d=a+(3âˆ’1)d

Fourth term is {a_4} = {a_3} + d = \left( {a + 2d} \right) + d = a + 3d = a + \left( {4 – 1} \right)da

4

â€‹

=a

3

â€‹

+d=(a+2d)+d=a+3d=a+(4âˆ’1)d

Looking at this pattern, we can say that nth term {a_n} = a + \left( {n – 1} \right)da

n

â€‹

=a+(nâˆ’1)d where a is the first term and d is the common difference of the arithmetic sequence.

{a_n}a

n

â€‹

is called the general term of the AP. Sometimes, it is also denoted by ll.

Applications of arithmetic sequence:

We use arithmetic sequence very often in our real life. When we ride a taxi, this also has an arithmetic sequence. When we take a taxi, we are charged an initial fixed rate and then a per kilometer charge is applied. This is an arithmetic sequence where for every kilometer you will be charged a certain constant rate plus the initial rate.

We use it in calculating our financing estimates, like simple and compound interest, loan amortization. Arithmetic sequence can be applied in almost all aspects of our lives. We just have to analyze how we use it in our day-to-day life. Having knowledge about this kind of sequence can give us a different perspective on how things happen in our lives.

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