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Solve the equation for $x$: $1+4+7+ 10 + \dots + x = 287$
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$1+4+7+10 + \dots + x = 287$
$a = 1, d = 3$ Last term $= x$
$\Rightarrow 1 + (n - 1) 3 = x$
$3n - 2 = x \Rightarrow n = \frac{x + 2}{3}$
dots (i)
$S_n = 287$
$\frac{n}{2} [1 + x] = 287 \Rightarrow \frac{(\frac{x + 2}{3})}{2} [1 + x] = 287$
quad using (i)
$\frac{(x + 2)(x + 1)}{6} = 287$
$(x + 2)(x + 1) = 1722$
$x^2 + 3x + 2 = 1722$
$x^2 + 3x - 1720 = 0$
$(x + 43)(x - 40) = 0$
$\Rightarrow x = -43, 40$
$x \ne -43$ (as terms are increasing positive numbers)
$\therefore x = 40$
$a = 1, d = 3$ Last term $= x$
$\Rightarrow 1 + (n - 1) 3 = x$
$3n - 2 = x \Rightarrow n = \frac{x + 2}{3}$
dots (i)
$S_n = 287$
$\frac{n}{2} [1 + x] = 287 \Rightarrow \frac{(\frac{x + 2}{3})}{2} [1 + x] = 287$
quad using (i)
$\frac{(x + 2)(x + 1)}{6} = 287$
$(x + 2)(x + 1) = 1722$
$x^2 + 3x + 2 = 1722$
$x^2 + 3x - 1720 = 0$
$(x + 43)(x - 40) = 0$
$\Rightarrow x = -43, 40$
$x \ne -43$ (as terms are increasing positive numbers)
$\therefore x = 40$