Question

Find the sum of the following arithmetic progressions:

3, 9/2, 6, 15/2, ... to 25 terms

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

3, 9/2, 6, 15/2, ... to 25 terms

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

`= 9/2 - 3`

`= (9 - 6)/2`

`= 3/2`

Number of terms (n) = 25

The first term for the given A.P. (a) = 3

So, using the formula we get,

`S_25 = 25/2 [2(3) +(25 - 1)(3/2)]`

`= (25/2)[6 + (25)(3/2)]`

`= (25/2)[6 + (72/2)]`

`= (25/2) [42]`

= 525

On further simplifying we get

`S_25 = 252`

Therefore the sum of first 25 term for the given A.P. is 525