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

Answer

N/a

Explanation

Let total number of terms be n.
First term of progression `a_(1)=25`
2nd term of progression `a_(2)=22`
`therefore " Common difference "d=a_(2)-a_(1)=22-25=-3`
`therefore" "116=(n)/(2)[2xx25+(n-1)xx(-3)]`
`therefore " "116=(n)/(2)[50-3n+3]`
`therefore " "116=(n)/(2)[53-3n]`
`therefore" "232=53n-3n^(2)`
`therefore" "3n^(2)-(24+29)n+232=0`
`therefore" "3n^(2)-24n-29n+232=0`
`therefore " "3n(n-8)-29(n-8)=0`
`rArr" "(n-8)(3n-29)=0`
`rArr" "n=8 or (29)/(3)`
But the number of terms can not be in fraction from which `n=(29)/(3)` is not acceptable.
`therefore "Number. of terms n=8"`
Then 8th (last) term of progression
`a_(8)=a_(1)+(n-1)d`
`=25+(8-1)xx(-3)`
`=25+7xx(-3)=25-21=4`
Therefore, last term of progression = 4