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Question
Find the sum of the following arithmetic progressions:
3, 9/2, 6, 15/2, ... to 25 terms
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
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
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