(B) Given,the $AP$ is $10, 6, 2, \ldots$Here,the first term $a = 10$ and the common difference $d = 6 - 10 = -4$.The formula for the sum of the first

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(B) Given,the $AP$ is $10, 6, 2, \ldots$Here,the first term $a = 10$ and the common difference $d = 6 - 10 = -4$.The formula for the sum of the first $n$ terms of an $AP$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.For $n = 16$:$S_{16} = \frac{16}{2}[2(10) + (16 - 1)(-4)]$$S_{16} = 8[20 + 15(-4)]$$S_{16} = 8[20 - 60]$$S_{16} = 8(-40) = -320$.