KVPY SA Exam 2014 Question Paper

Let maximum house is 'n' ; sum of first 'n' even natural numbers $={\mathit{n}}^2+\mathit{n}$
Let first 'm' even natural numbers are left in numbering the houses.
$\left({\mathit{n}}^2+\mathit{n}\right)-\left({\mathit{m}}^2+\mathit{m}\right)=170$
$\Rightarrow {\mathit{n}}^2-{\mathit{m}}^2+\mathit{n}-\mathit{m}=170$
$\Rightarrow \left(\mathit{n}-\mathit{m}\right)\left(\mathit{n}+\mathit{m}+1\right)=170$
$\mathit{n}-\mathit{m}=10\Rightarrow \mathit{n}-\mathit{m}=10$
$\mathit{n}+\mathit{m}+1=17\Rightarrow \mathit{n}+\mathit{m}=16$
$\mathit{n}=13;\mathit{m}=3$
$\mathit{n}\le 13$
If first term is a – 10 then sixth = a
$\frac{\mathit{n}}{2}\left[2\left(\mathit{a}-10\right)+2\left(\mathit{n}-1\right)\right]=170$
$\Rightarrow \mathit{n}\left[\mathit{a}+\mathit{n}-11\right]=170$
$\Rightarrow \mathit{a}=\frac{170}{\mathit{n}}+11-\mathit{n}$
∴ n = 10 (only will make 'a' integer)
$\Rightarrow \mathit{a}=17+11-10=18$