# Compute: 2-4+6-8+10-12+14-...+210

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`2-4+6-8+10-12+14-` `...+210`

To compute this, group the positive numbers and group the negative numbers together. In getting the sum of the negative numbers, consider the last negative term.

`= 2 - 4 + 6 -8 + 10 - 12 + 14 -`` ... -208+210`

`=(2+6+10+14+...+210)+(-4-8-12-`` ...-208)`

Notice that both groups form an arithmetic series.

For the first group

`(2+6+10+14+...+210)`

its first term is 2 and its common difference is 4. To get the number of terms present in the first group, apply the formula:

`a_n =a_1+(n-1)d`

`210=2+(n-1)(4)`

`208=(n-1)(4)`

`52=n-1`

`53=n`

So there are 53 terms in the first group.

Then, compute the sum of the first group by applying the formula of finite arithmetic series which is:

`S_n = n/2(a_1+a_n)`

`S_53=53/2(a_1+a_53)=53/2(2+210)=5618`

For the second group

`(-4-8-12-` `...-208)`

its first term is -4 and its common difference is -4. To get the number of terms present in the second group, apply the formula:

`a_n=a_1+(n-1)d`

`-208=-4+(n-1)(-4)`

`-204=(n-1)(-4)`

`51=n-1`

`52=n`

So, there are 52 terms in the second group.

To compute the sum, apply the formula of finite arithmetic series.

`S_n=n/2(a_1+a_n)`

`S_52=52/2(a_1+a_52)=52/2(-4+(-208))=-5512`

Adding the results of the two groups, yield:

`(2+6+10+14+...+210) + (-4-8-12-` `...-208)`

`=5618 + (-5512)`

`=106`

**Therefore, `2-4+6-8+10-12+14-` `...+210 = 106` .**