Find in terms of 𝑛 the sum of the
arithmetic sequence nine, 10, 11, and so on up to 𝑛 plus eight.
We know that we can calculate the
sum of any arithmetic sequence using one of two formulae. Firstly, 𝑠 sub 𝑛 is equal to 𝑛
over two multiplied by two 𝑎 plus 𝑛 minus one multiplied by 𝑑. In this formula, 𝑎 is the first
term and 𝑑 is the common difference. Alternatively, 𝑠 sub 𝑛 is equal
to 𝑛 over two multiplied by 𝑎 plus 𝑙, where 𝑎 is the first term and 𝑙 is the
We can see from our sequence that
the first term 𝑎 is nine and the common difference is one as the numbers are
increasing by one. The value of our last term 𝑙 is 𝑛
plus eight. Substituting in our values of 𝑎
and 𝑑 to the first formula gives us 𝑛 over two multiplied by two multiplied by
nine plus 𝑛 minus one multiplied by one. This simplifies to 𝑛 over two
multiplied by 18 plus 𝑛 minus one. We can simplify the square bracket
further such that the sum of the first 𝑛 terms is equal to 𝑛 over two multiplied
by 𝑛 plus 17.
If we chose to use the second
formula, we need to substitute 𝑎 equals nine and 𝑙 equals 𝑛 plus eight. This gives us 𝑠 sub 𝑛 is equal to
𝑛 over two multiplied by nine plus 𝑛 plus eight. Once again, the bracket or
parentheses simplifies to 𝑛 plus 17.
The sum of the arithmetic sequence
in terms of 𝑛 is 𝑛 over two multiplied by 𝑛 plus 17.