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`


Discussion

No Comment Found

Related InterviewSolutions