Question

Find the sum 2 + 4 + 6 ... + 200

Answer 1

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

`S_n =  n/2[2a + (n -1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

2 + 4 + 6 ... + 200

Common difference of the A.P. (d)  = `a_2 - a_1`

= 6 - 4

= 2

So here,

First term (a) = 2

Last term (l) = 200

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

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

So, for the last term,

200 = 2 +(n - 1)2

200 = 2 + 2n - 2

200 = 2n

Further simplifying,

`n = 200/2`

n = 100

Now, using the formula for the sum of n terms, we get

`S_n = 100/2 [2(2) + (100 - 1)2]`

= 50 [4 + (99)2]

= 50(4 + 198)

On further simplification we get

`S_n = 50(202)`

= 10100

Therefore, the sum of the A.P is `S_n = 10100`