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(A) Let the sum of $n$ terms of the given $A.P.$ be $116$.$S_{n} = \frac{n}{2}[2a + (n - 1)d]$Here,$a = 25$ and $d = 22 - 25 = -3$.$	herefore 116 = \

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$4$

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(A) Let the sum of $n$ terms of the given $A.P.$ be $116$.$S_{n} = \frac{n}{2}[2a + (n - 1)d]$Here,$a = 25$ and $d = 22 - 25 = -3$.$	herefore 116 = \frac{n}{2}[2 	imes 25 + (n - 1)(-3)]$$\Rightarrow 232 = n(50 - 3n + 3)$$\Rightarrow 232 = 53n - 3n^{2}$$\Rightarrow 3n^{2} - 53n + 232 = 0$$\Rightarrow 3n^{2} - 24n - 29n + 232 = 0$$\Rightarrow 3n(n - 8) - 29(n - 8) = 0$$\Rightarrow (n - 8)(3n - 29) = 0$Since $n$ must be a positive integer,$n = 8$.$	herefore a_{8} = a + (8 - 1)d = 25 + 7(-3) = 25 - 21 = 4$.Thus,the last term of the $A.P.$ is $4$.

If the sum of a certain number of terms of the $A.P.$ $25, 22, 19, \ldots$ is $116$,find the last term.

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