Question

Find the sum:

25 + 28 + 31 + ... + 100

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

25 + 28 + 31 + ... + 100

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

= 28 - 25

= 3

So here,

First term (a) = 25

Last term (`l`) = 100

Common difference (d) = 3

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,

100 = 25 + (n - 1)(3)

100 = 25 + 3n - 3 

100 = 22 + 3n

100 - 22 = 3n

Further solving for n,

78 = 3n

`n = 78/3`

n = 26

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

`S_n = 26/2[2(25) = (26 - 1)(3)]`

= 13[50 + (25)(3)]

= 13(50 + 75)

= 13(125)

On further simplification, we get,

`S_n =  1625`

Therefore, the sum of the A.P is `S_n = 1625`