Question

Find x, if −4 + (−1) + 2 + ... + x = 437.

Answer 1

Here, a = -4 and d = [−1 − (−4)] = 3
Let x be the n^th term of the AP.
Sum of n terms of an AP is given by

$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ \Rightarrow \frac{n}{2}\left[2\times (-4)+\left(n-1\right)\times 3\right]=437 \\ \Rightarrow \frac{n}{2}\left[-8+3n-3\right]=437 \\ \Rightarrow n[3n-11]=437\times 2 \\ \Rightarrow 3n^2-11n-874=0 \\ \Rightarrow 3n^2-57n+46n-874=0 \\ \Rightarrow 3n(n-19)+46(n-19)=0 \\ \Rightarrow (3n+46)(n-19)=0 \\ \Rightarrow n=19orn=\frac{-46}{3}(n=\frac{-46}{3}\text{is rejected, as}n\text{can't be a fraction}) \end{gathered}$$
Thus, the sum of 19 terms is equal to 437.
∴ x = T₁₉​ = a + (n −1)d
⇒ a + 18d
⇒ (−4) + 18 ⨯ 3 = 50

Hence, x = 50