If the sum of a certain number of terms of the A.P. 25, 22, 19.... is 116. Find the last term.

Explanation

Here, first term, `a = 25`

Common difference, `d = -3`

Sum,`S_n = 116`

So,

`n/2(2a+(n-1)d ) = 116`

`=>n/2(2**25+(n-1)(-3) ) = 116`

`=>n(50+3-3n) = 232`

`=>3n^2-53n +232 = 0`

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