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प्रश्न
Find the sum 25 + 28 + 31 + ….. + 100
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उत्तर
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`
