# Given the First and Last Terms, How to Form the Sum of the First n Terms of every Arithmetic Series?

Arithmetic series
The formula is Snn[2a+(n-1)d] where Sn is the sum of n terms, a is the first term, n is the number of terms and d is the common difference.

Proof
The general term of an arithmetic series is an=a+(n-1)d.

Sn=a1+a2+⋯+a(n-1)+an
Sn=a+(a+d)+⋯+[a+(n-2)d]+[a+(n-1)d]…(i)

If we write the series in reverse we get:

Sn=[a+(n-1)d]+[a+(n-2)d]+⋯+(a+d)+a … (ii)

We can add (i) and (ii).

2Sn=[2a+(n-1)d]+[2a+(n-1)d]+⋯+[2a+(n-1)d]+[2a+(n-1)d]
2Sn=n[2a+(n-1)d]
Snn[2a+(n-1)d]
Alternative Proof

Sn=a+[a+d]+[a+2d]+…+[d]+ … (i)

In reverse

Sn=+[l-d]+[l-2d]+…+[a+d]+a …(ii)

2Sn=[a+]+[a+]+⋯+[a+]… n times
2Sn=n[a+]
Snn(a+)

Arithmetic Series:

The sum of the terms of an Arithmetic sequence is called as Arithmetic series. For example:

7, 17, 27, 37, 47, … is an A.P.

7+17+27+37+47+… is Arithmetic series.
The sum of n terms of an Arithmetic Sequence:

The general form of an arithmetic sequence is a, a+d, a+2d, …, a+(n-1)d

Let Sn denoted the sum of n terms of an Arithmetic sequence.
Then Sn=a+(a+d)+(a+2d)…+[a+(n-1)d]
Let nth term [a+(n-1)d]=

The above series can be written as

Sn=a+(a+d)+(a+2d)+⋯+
Or, Sn=a+(a+d)+(a+2d)+⋯+(-2d)+(d+ … (I)

Writing (I) in reverse order, we have

Sn=+(d+(-2d)+⋯+(a+2d)+(a+d)+a … (II)

2Sn=(a+)+(a+)+(a+)+⋯+(a+)
2Sn=n(a+)
Snn(a+), but =a+(n-1)d
Snn[a+a+(n-1)d]
Snn[2a+(n-1)d]

is the formula for the sum of n terms of an arithmetic sequence.

The sum of an arithmetic series
Sometimes we want to add the terms of a sequence. What would we get if we wanted to add the first n terms of an arithmetic progression? We would get

Sn=a+(a+d)+(a+2d)+⋯+(-2d)+(d+.

Now this is now a series, as we have added together the n terms of a sequence. This is an arithmetic series, and we can find its sum by using a trick. Let us write the series down again, but this time we shall write it down with the terms in reverse order. We get

Sn=+(d+(-2d)+⋯+(a+2d)+(a+d)+a.

We are now going to add these two series together. On the left-hand side, wejust get 2Sn. But on the right-hand side, we are going to add the terms in the two series so that each term in the first series will be added to the term vertically below it in the second series. We get

Sn=+(a+)+(a+)+⋯+(a+)+(a+)+(a+),

and on the right-hand side there are n copies of (a+) so we get

2Sn=n(a+)

But of course we want Sn rather than 2Sn, and so we divide by 2 to get

Snn(a+)

We have found the sum of an arithmetic progression in terms of its first and last terms, a and , and the number of terms n.

We can also find an expression for the sum in terms of the a, n and the common difference d. To do this, we just substitute our formula for into our formula for Sn. From

=a+(n-1)d, Snn(a+)

we obtain

Snn(a+a+(n-1)d)
n(2a+(n-1)d).

Key Point
The sum of the terms of an arithmetic progression gives an arithmetic series. If the starting value is a and the common difference is d then the sum of the first n terms is

Snn(2a+(n-1)d).

If we know the value of the last term instead of the common difference d then we can write the sum as

Snn(a+)

Partial Sums of Arithmetic Sequences
There is a charming story told about Carl Freidrich Gauss, one of the greatest mathematicians of all time. Early in Gauss’ school career, the schoolmaster as- signed the class the task of summing the first hundred positive integers, 1+2+3+…+99+100. That should have occupied a good portion of the morning, but while other class members busied themselves at their slates calculating 1+2=3, 3+3=6, 6+4=10, and so on, Gauss sat quietly for a few moments, wrote a single number on his slate, and presented it to the teacher. Young Gauss observed that 1 and 100 add up to 101, as do the pair 2 and 99, 3 and 98, and so on up to 50 and 51. There are fifty such pairs, each with a sum of 101, for a total of 50⋅101=5050, the number he wrote on his slate.

This approach works for the partial sum of any arithmetic sequence, and we will use the method to derive some useful formulas. However, the ideas are more valuable than memorizing formulas. If you understand the idea, you can recreate the formula when needed.

To find a formula for the nth partial sum of an arithmetic sequence, that is, the sum of n consecutive terms, pair the first and last terms, the second and next-to-last, and so on; each pair has the same sum. In fact, it is easier to pair all terms twice, as illustrated with Gauss’ sum: The sum on the right has 100 terms, so 2S100=100(101). Dividing by 2, S100=½⋅100(1+100)=5050.

For the general case, pairing the terms in Sn and adding gives 2Sn=n(a1+an) because there are n pairs, each with the same sum. Dividing by 2 yields the desired formula.

Partial sums of an arithmetic sequence
Suppose {an} is an arithmetic sequence. The sum of the first n terms is given by

Snn(a1+an)

The formula is probably most easily remembered as n times the average of the first and last terms.

Sum of First n Terms of an AP
Let us consider the situation: In which Renata put \$100 into her daughter’s money box when she was one year old, \$150 on her second birthday, \$200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old? Here, the amount of money (in \$) put in the money box on her first, second, third, fourth … birthday were respectively 100, 150, 200, 250, … till her 21st birthday. To find the total amount in the money box on her 21 st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.

We consider the problem given to Gauss, to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote:

S=1+2+3+…+99+100

And then, reversed the numbers to write

S=100+99+…+3+2+1

2S=(100+1)+(99+2)+…+(3+98)+(2+99)+(1+100)
=101+101+…+101+101 (100 times)
S=(100×101)/2=5050

Thus, the sum =5050.

We will now use the same technique to find the sum of the first n terms of an AP:

a, a+d, a+2d, …

The nth term of this AP is a+(n-1)d. Let S denote the sum of the first n terms of the AP. We have

Sn=a+(a+d)+(a+2d)+…+[a+(n-1)d]…(1)

Rewriting the terms in reverse order, we have

Sn=[a+(n-1)d]+[a+(n-2)d]+…+(a+d)+a … (2)

On adding (1) and (2), term-wise. we get

2Sn=[2a+(n -1)d]+[2a+(n-1)d]+…+[2a+(n -1)d]
n times
2Sn=n[2a+(n -1)d]…(Since, there are n terms)
Snn[2a+(n -1)d]

So, the sum of the first n terms of an AP is given by

Snn[2a+(n-1)d]

We can also write this as Snn[a+a+(n-1)d]

Snn[a+an ]…(3)

Now, if there are only n terms in an AP, then an=, the last term. From (3), we see that

Snn[a+]…(4)

This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.

Now we return to the question that was posed to us in the beginning. The amount of money (in \$) in the money box of Renata’s daughter on 1st, 2nd, 3rd, 4th birthday, …, were 100, 150, 200, 250, …, respectively.

This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a=100, d=50 and n=21. Using the formula:

Snn[2a+(n-1)d],

we have S21=½⋅21⋅[2⋅100+(21-1)⋅50]=½⋅21⋅[200+1000]

=½⋅21⋅1200=12600

So, the amount of money collected on her 21st birthday is \$12600.

Hasn’t the use of the formula made it much easier to solve the problem?

We use Sn to denote the sum of first n terms of the AP. We write S20 to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities Sn, a, d and n. If we know any three of them, we can find the fourth.

Remark: The nth term of an AP is the difference of the sum to first It terms and the sum to first (n-1) terms of it, i.e., an=SnS(n-1).

Let us consider some examples in Get Terms from Sequences.

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