## SAT Math Concepts : Arithmetic Sequence

A sequence of numbers in which the next term is obtained by adding a constant ‘d’ to the previous term is known as an arithmetic sequence or arithmetic progression.

The following sequence is an arithmetic sequence : 4, 9, 14, 19, 24.

The first term of this sequence is 4 and subsequent terms are obtained by adding 5 to the previous term.

The constant ‘d’ that is added to each term is know as the common difference of the Arithmetic Sequence. It is called the common difference as the difference between any two consecutive term of the sequence is the same.

So, an arithmetic sequence may also be defined as one in which the difference between any two consecutive terms is the same.

### Finding the nth term of an Arithmetic Sequence

Let us take a look at the sequence again. 4, 9, 14, 19, 24

The second term 9 = 4 + 5 (first term + one common difference)

The third term 14 = 4 + 5 + 5 or 14 = 4 + 2 * 5 (first term + two common differences)

The fourth term 19 = 4 + 5 + 5 + 5 = 4 + 3 * 5 (first term + three common differences)

Extending the same logic, the nth term of the sequence is the firs term + (n – 1) common differences.

Or tn, the nth term = t1 + (n – 1)d, where t1 is the first term and d is the common difference.

### Illustrative Example

What is the 15th term of an arithmetic sequence if its first term is 21 and its common difference is 8?

t15 = t1 + (15 – 1)d

t1 = 21 and d = 8

So, t15 = 21 + 14 * 8

or t15 = 21 + 112 = 133

### Finding the sum of first n terms of an Arithmetic Progression

Let us compute the sum using brute force. The sum of 4, 9, 14, 19 and 24 is 70.

The average of the 5 numbers is $70 \over 5 \\$ = 14.

For any set of ‘n’ numbers, whose average is ‘a’, the sum of those ‘n’ numbers will be n * a.

So, if we can determine the average of the ‘n’ terms of an arithmetic sequence, say a, its sum will be n * a.

In the example we computed the average to be 14.

Notice that 14 happens to be the middle term of the arithmetic sequence.

Note : For an arithmetic sequence, the middle is always the average of the sequence.

So, the sum of an arithmetic sequence is the product of its middle term and the number of terms.
i.e., sum = middle term * number of terms.

The middle term of the sequence being the average is no strange coincidence.

If the 5 terms of the sequence are written as 14 – 2d, 14 – d, 14, 14 + d and 14 + 2d, it becomes evident that when summing the 5 numbers, the -2d and +2d of the first and fifth term will cancel out. So, will the -d and +d of the second and the fourth term.

Therefore, the sum of the 5 terms will be 5 * 14 = 70.

As the common differences of terms that are equidistant from the middle terms cancel out, what we will be left with is n times the middle term.

Hence, the middle term will always be the average of an arithmetic sequence.

### What if there are even number of terms in the sequence?

If there are odd number of terms, the middle term will be the arithmetic mean or the average.

If there are even number of terms, there is no middle term. The average of the sequence will be the arithmetic mean of the middle two terms.

The average of an arithmetic sequence, whether it has an odd number of terms or an even number of terms, is the average of two terms that are equidistant from the middle term.

In the sequence, 4, 9, 14, 19, 24 – the average 14 is the average of 9 and 19. It is also the average of 4 and 24.

The first and the last term of the sequence are equidistant from the middle term.

So, we can express the sum of an arithmetic sequence as $\lbrack{ {{first \; term \;+ \;last \; term} \over 2} \rbrack} * number \; of \; terms\\$.

The last term $t_n = t_1 + \lbrack {n – 1} \rbrack d \\$, where $t_1 \\$ is the first term.

Therefore, first term + last term = ${t_1 + t_n} \\$ = $t_1 + t_1 + \lbrack n – 1 \rbrack d = \ 2 t_1 + \lbrack n – 1 \rbrack d \\$.

Hence, the sum of an arithmetic sequence may also be expressed as $n \over 2 \\$ $\lbrack{ 2t_1 + \lbrack n - 1 \rbrack d \rbrack} \\$.