What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?
-30
Step 1 — Find a and d:
From \(a_3 = a+2d = -13\) and \(a_6 = a+5d = -4\), subtract to get \(3d = 9 \implies d = 3\), so \(a = -13 - 6 = -19\).
Step 2 — Apply sum formula:
\[S_{12} = \frac{12}{2}\big[2(-19) + 11(3)\big] = 6[-38+33] = 6(-5)\]
Therefore \(S_{12} = \mathbf{-30}\).
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