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Question
If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
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Solution 1
Let the sum of n terms of the given A.P. be 116.
Solution 2
In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the last term for that A.P.
So here, let us first find the number of terms whose sum is 116. For that, we will use the formula,
`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
So for the given A.P (25, 22, 19, .....)
The first term (a) = 25
The sum of n terms `S_n = 116`
Common difference of the A.P (d) = `a_2 - a_1`
= 22 - 25
= -3
So, on substituting the values in the formula for the sum of n terms of A.P. we get
`116 = n/2 [2(25) + (n - 1)(-3)]`
`116 = (n/2)[50 + (-3n + 3)]`
`116 = (n/2) [53 - 3n]`
`(116)(2) = 53n - 3n^2`
So, we get the following quadratic equation,
`3n^2 - 53n + 232 = 0`
On solving by splitting the middle term,we get
`3n^2 - 24n - 29n + 232 = 0`
3n(n - 8) - 29(n - 8) = 0
(3n - 29)(n - 8) = 0
Further
3n - 29 = 0
`n = 29/3`
Also
n - 8 = 0
Now since n cannot be a fraction, so the number of terms is a8
`a_8 = a_1 + 7d`
= 25 + 7(-3)
= 25 - 21
= 4
Therefore the last term of the given A.P such that the sum of terms is 116 is 4
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