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An AP consists of '$n$' terms whose $n^{\text{th}}$ term is $4$ and the common difference is $2$. If the sum of '$n$' terms of AP is $-14$, then find '$n$'. Also, find the sum of the first $20$ terms.
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Let first term = $a$, common difference = $$\begin{aligned}& d = 2 \\ & a_n = a + (n - 1)2 = 4 \\ & text{ATQ, } a + 2n = 6 \\ & a = 6-2n \\ & text{ATQ, } S_n = \frac{n}{2}[2a + (n - 1)2] = -14 \\ & n[a + n - 1] = -14 \\ & n[6 - 2n + n - 1] = -14 \\ & n^2 - 5n - 14 = 0 \\ & Rightarrow n = 7 \\ & text{and } a = -8 \\ & S_{20} = \frac{20}{2}[2\times (-8) + 19 \times 2] \\ & = 220\end{aligned}$$