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

6. 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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Given : A.P. 25, 22, 19, ….....

S_n=116

a=25  , d = -3

S_n=\frac{n}{2}[2a+(n-1)d]

\Rightarrow \, \, 116=\frac{n}{2}[2(25)+(n-1)(-3)]

\Rightarrow \, \, 232=n[50-3n+3]

\Rightarrow \, \, 232=n[53-3n]

\Rightarrow \, \, 3n^2-53n+232=0

\Rightarrow \, \, 3n^2-24n-29n+232=0

\Rightarrow \, \, 3n(n-8)-29(n-8)=0

\Rightarrow \, \, (3n-29)(n-8)=0

\Rightarrow \, \, n=8\, \, or\, \, \, n=\frac{29}{3}

n could not be \frac{29}{3} so n=8.

Last term =a_8=a+(n-1)d

                        =25+(8-1)(-3)

                      =25-21=4

The, last term of A.P. is 4.

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