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

#### Solution

The given A.P. is 25,.22,.19.....

Here, a = 25, d = \[22 - 25 = - 3\]

\[S_n = 116\]

\[ \Rightarrow \frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right] = 116\]

\[ \Rightarrow n\left[ 2 \times 25 + \left( n - 1 \right)\left( - 3 \right) \right] = 232\]

\[ \Rightarrow 50n - 3 n^2 + 3n = 232\]

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

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

\[ \Rightarrow n(3n - 29) - 8(3n - 29) = 0\]

\[ \Rightarrow (3n - 29)(n - 8) = 0\]

\[ \Rightarrow n = \frac{29}{3} \text { or }8\]

\[\text { Since n cannot be a fraction, } n = 8 . \]

\[\text {Thus, the last term }: \]

\[ a_n = a + (n - 1)d\]

\[ \Rightarrow a_8 = 25 + \left( 8 - 1 \right) \times \left( - 3 \right)\]

\[ \Rightarrow a_8 = 4\]