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Question
Find the sum 2 + 4 + 6 ... + 200
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Solution
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`
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