

InterviewSolution
Saved Bookmarks
1. |
If the sum of a certain number of terms of the A.P. 25, 22, 19.... is 116. Find the last term. |
Answer» 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` `=>3n^2-29n-24n+232 = 0` `=>n(3n-29)-8(3n-29) = 0` `=>(n-8)(3n-29) = 0` `=>n = 8 and n = 29/3` As n is a whole number. So, it can not be `29/3` `:.` Number of terms,`n = 8`. `:.` Last term `= a+(n-1)d = 25+7(-3) = 4` |
|