# Find the sum of a given arithmetic sequence: sum n=1 100 (6 - 12 n)

Polynomial arithmetic
Find the sum of a given arithmetic sequence:
$$\displaystyle\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}\ {n}\right)}$$

2021-01-28
Formula used:
a) $$\displaystyle\sum\ {k}={1}\ {n}{k}={n}\ {2}\ {\left({n}\ +\ {1}\right)}$$
b) $$\displaystyle\sum\ {k}={1}\ {n}\ {c}={c}{n}$$
Calculation:
In order to find the sum of an arithmetic sequence, you need
$$\displaystyle{S}\ {n}=\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}{n}\right)}$$
Write $$\displaystyle\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}{n}\right)}$$ as follows,
$$\displaystyle\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}{n}\right)}=\sum\ {n}={1}\ {100}\ {6}\ -\ {12}\ \sum\ {n}={1}\ {100}\ {n}$$
Now to find $$\displaystyle\sum\ {n}={1}\ {100}\ {6},$$
Substitute the values of n in equation (b).
We get,
$$\displaystyle\sum\ {n}={1}\ {100}\ {6}$$
$$\displaystyle={6}\ {\left({100}\right)}$$
$$\displaystyle={600}$$
Now to find the $$\displaystyle{12}\ \sum\ {n}={1}\ {100}\ {n}$$
Substitute the values of n in equation (a).
We get,
$$\displaystyle{12}\ \sum\ {n}={1}\ {100}\ {n}$$
$$\displaystyle={12}\ {\left({100}{\left({100}\ +\ {1}\right)}{2}\right)}$$
$$\displaystyle={\left({100}\ \times\ {101}\ {4}\right)}$$
$$\displaystyle={2525}$$
Then,
$$\displaystyle\sum\ {n}={1}\ {100}\ {6}\ -\ {12}\ \sum\ {n}={1}\ {100}\ {n}={600}\ -\ {2525}$$
$$\displaystyle=\ -{1925}$$
Hence
$$\displaystyle\sum\ {n}={1}\ {100}{\left({6}\ -\ {12}{n}\right)}\ =\ -{1925}$$