70
Solve the equation for $x : - 4 + (-1) + 2 + \ldots + x = 437$.
Show SolutionHide Solution↓
This is an Arithmetic Progression (AP).
$a = -4$, $d = -1 - (-4) = 3$.
Let $x$ be the $n^{th}$ term, so $a_n = x$.
Sum of $n$ terms $S_n = 437$.
$S_n = \frac{n}{2}[2a + (n-1)d]$
$437 = \frac{n}{2}[2(-4) + (n-1)3]$
$437 = \frac{n}{2}[-8 + 3n - 3]$
$437 = \frac{n}{2}[3n - 11]$
$874 = 3n^2 - 11n$
$3n^2 - 11n - 874 = 0$
Using quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $n = \frac{11 \pm \sqrt{(-11)^2 - 4(3)(-874)}}{2(3)}$
$n = \frac{11 \pm \sqrt{121 + 10488}}{6} = \frac{11 \pm \sqrt{10609}}{6} = \frac{11 \pm 103}{6}$
$n = \frac{11+103}{6} = \frac{114}{6} = 19$ or $n = \frac{11-103}{6} = \frac{-92}{6}$ (not possible as $n$ must be positive integer).
So, $n = 19$.
Now find $x = a_n = a_{19}$.
$a_n = a + (n-1)d$
$x = -4 + (19-1)3 = -4 + 18(3) = -4 + 54 = 50$.
So, $x = 50$.
$a = -4$, $d = -1 - (-4) = 3$.
Let $x$ be the $n^{th}$ term, so $a_n = x$.
Sum of $n$ terms $S_n = 437$.
$S_n = \frac{n}{2}[2a + (n-1)d]$
$437 = \frac{n}{2}[2(-4) + (n-1)3]$
$437 = \frac{n}{2}[-8 + 3n - 3]$
$437 = \frac{n}{2}[3n - 11]$
$874 = 3n^2 - 11n$
$3n^2 - 11n - 874 = 0$
Using quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $n = \frac{11 \pm \sqrt{(-11)^2 - 4(3)(-874)}}{2(3)}$
$n = \frac{11 \pm \sqrt{121 + 10488}}{6} = \frac{11 \pm \sqrt{10609}}{6} = \frac{11 \pm 103}{6}$
$n = \frac{11+103}{6} = \frac{114}{6} = 19$ or $n = \frac{11-103}{6} = \frac{-92}{6}$ (not possible as $n$ must be positive integer).
So, $n = 19$.
Now find $x = a_n = a_{19}$.
$a_n = a + (n-1)d$
$x = -4 + (19-1)3 = -4 + 18(3) = -4 + 54 = 50$.
So, $x = 50$.