Question

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

Answer 1

The given A.P. is 25,.22,.19.....
Here, a = 25, d = $22-25=-3$
$$\begin{gathered} S_n=116 \\ \Rightarrow \frac{n}{2}\left[2a+\left(n-1\right)d\right]=116 \\ \Rightarrow n\left[2\times 25+\left(n-1\right)\left(-3\right)\right]=232 \\ \Rightarrow 50n-3n^2+3n=232 \\ \Rightarrow 3n^2-53n+232=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow 3n^2-29n-24n+232=0 \\ \Rightarrow n(3n-29)-8(3n-29)=0 \\ \Rightarrow (3n-29)(n-8)=0 \\ \Rightarrow n=\frac{29}{3}or8 \\ \mathrm{Since}n\mathrm{cannot}\mathrm{be}\mathrm{a}\mathrm{fraction},n=8. \\ \mathrm{Thus},\mathrm{the}\mathrm{last}\mathrm{term}: \\ {a}_n=a+(n-1)d \\ \Rightarrow a_8=25+\left(8-1\right)\times \left(-3\right) \\ \Rightarrow a_8=4 \end{gathered}$$