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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Solution

The given A.P. is 25, 22, 19, …..
Here, a = 25, d = 22 - 25 = -3
$S_{n} = 116 \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 116 \Rightarrow n [2 \times 25 + (n - 1) (- 3)] = 232 \Rightarrow 50 n - 3 n^{2} + 3 n = 232 \Rightarrow 3 n^{2} - 53 n + 232 = 0 \Rightarrow 3 n^{2} - 29 n - 24 n + 232 = 0 \Rightarrow n (3 n - 29) - 8 (3 n - 29) = 0 \Rightarrow (3 n - 29) (n - 8) = 0 \Rightarrow n = \frac{29}{3}$ or $8$
Since n cannot be a friction, n = 8
Thus, the last term:
$a_{n} = a + (n - 1) d \Rightarrow a_{8} = 25 + (8 - 1) \times (- 3) \Rightarrow a_{8} = 4$

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