Question

The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is $170$. If there are at least $6$ houses in that row and $a$ is the number of the sixth house, then

• A. $2 \le a \le 6$
• B. $8 \le a \le 12$
• C. $14 \le a \le 20$
• D. $22 \le a \le 30$

Answer 1

Answer: (C)

Let the number of houses be
$x, x + 2, x + 4, x + 6, x + 8, x + 10,...$
$6$th number of house is $a$.
$\because x + 10 = a$
$\Rightarrow x = a - 10$
$\therefore x > 10$
Now, $S_n = \frac{n}{2}(2x + (n - 1) 2)$
$S_n = n(x + n - 1)$
$\Rightarrow 170 = n(a - 10 + n - 1)$
$\Rightarrow n^2 + (a - 11)n - 170 = 0$
$\Rightarrow n = -\frac{\left(a-11\right)\pm\sqrt{\left(a - 11\right)^{2}+680}}{2} $
$\Rightarrow n = \frac{\left(11 - a\right)\pm\sqrt{\left(a - 11\right)^{2}+680}}{2} n\ge 6 $
$\because \frac{\left(11-a\right)\pm \sqrt{\left(a - 11\right)^{2}+680}}{2} \ge 6 $
$\Rightarrow a \le \frac{800}{24} \le 33.33 $
$\because 12 \le a \le 32$
$ a = 12, 14, 16, 18,... $
When, $a = 18, n = 10$, then $S_{n} = 170 $
$ \because a = 18$