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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
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The correct answer is:
$14 \leq a \leq 20$
Let house no are $\alpha, \alpha,+2, \alpha+4, \alpha+6, \alpha+8, \alpha+10, \ldots .$ $\alpha+10=a \Rightarrow \alpha=a-10\quad ......\text {(1)}$
House no. will be $(+)$
$\begin{array}{l}
\Rightarrow \alpha=a-10>0 \\
\Rightarrow \alpha>10 \\
\Rightarrow \alpha \geq 12 \text { as a is each too }\quad ......\text {(2)}
\end{array}$
Now $S_{n}=\frac{n}{2}[2 \alpha+(n-1) d]$
$\begin{array}{l}
170=\frac{n}{2}[2 \alpha+(n-1)(2)] \\
=n(\alpha+(n-1)) \\
=n(a-10+n-1) \\
=n(a-11+n) \\
\Rightarrow n^{2}+n(a-11)-170=0 \\
\Rightarrow n=\frac{(11-a) \pm \sqrt{(a-11)^{2}+680}}{2} \quad ......\text {(3)}\\
\because n \geq 6 \\
\Rightarrow \frac{(11-a) \pm \sqrt{(a-11)^{2}+680}}{2} \geq 6 \\
\Rightarrow a \leq \frac{800}{24}\quad ......\text {(4)}
\end{array}$
From $(2)$ and $(4) \Rightarrow 12 \leq a \leq 32$
Now checking through $(3)$ for $a=12,14, \ldots . .$;
we have $a=18, n=10$ and $S_{n}=170$
House no. will be $(+)$
$\begin{array}{l}
\Rightarrow \alpha=a-10>0 \\
\Rightarrow \alpha>10 \\
\Rightarrow \alpha \geq 12 \text { as a is each too }\quad ......\text {(2)}
\end{array}$
Now $S_{n}=\frac{n}{2}[2 \alpha+(n-1) d]$
$\begin{array}{l}
170=\frac{n}{2}[2 \alpha+(n-1)(2)] \\
=n(\alpha+(n-1)) \\
=n(a-10+n-1) \\
=n(a-11+n) \\
\Rightarrow n^{2}+n(a-11)-170=0 \\
\Rightarrow n=\frac{(11-a) \pm \sqrt{(a-11)^{2}+680}}{2} \quad ......\text {(3)}\\
\because n \geq 6 \\
\Rightarrow \frac{(11-a) \pm \sqrt{(a-11)^{2}+680}}{2} \geq 6 \\
\Rightarrow a \leq \frac{800}{24}\quad ......\text {(4)}
\end{array}$
From $(2)$ and $(4) \Rightarrow 12 \leq a \leq 32$
Now checking through $(3)$ for $a=12,14, \ldots . .$;
we have $a=18, n=10$ and $S_{n}=170$
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