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(C) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} \{2a + (n-1)d\}$.For $S_4 = 16$,we have $\frac{4}{2} \{2a + 3d\} = 16$,

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$-320$

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(C) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} \{2a + (n-1)d\}$.For $S_4 = 16$,we have $\frac{4}{2} \{2a + 3d\} = 16$,which simplifies to $2a + 3d = 8$ (Equation $1$).For $S_6 = -48$,we have $\frac{6}{2} \{2a + 5d\} = -48$,which simplifies to $2a + 5d = -16$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(2a + 5d) - (2a + 3d) = -16 - 8$,so $2d = -24$,which gives $d = -12$.Substituting $d = -12$ into Equation $1$: $2a + 3(-12) = 8$,so $2a - 36 = 8$,which gives $2a = 44$,so $a = 22$.Now,$S_{10} = \frac{10}{2} \{2a + 9d\} = 5 \{2(22) + 9(-12)\} = 5 \{44 - 108\} = 5 \{-64\} = -320$.

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

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