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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 $\frac{4}{2} [2a + 3d] = 16$,whic

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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 $3(2a + 5d) = -48$,or $2a + 5d = -16$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(2a + 5d) - (2a + 3d) = -16 - 8$,which gives $2d = -24$,so $d = -12$.Substituting $d = -12$ into Equation $1$: $2a + 3(-12) = 8$,so $2a - 36 = 8$,which means $2a = 44$,so $a = 22$.Now,$S_{10} = \frac{10}{2} [2a + 9d] = 5 [2(22) + 9(-12)]$.$S_{10} = 5 [44 - 108] = 5 [-64] = -320$.

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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