Question

If $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right].......\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$, then $n=$

• A. $27$
• B. $26$
• C. $376$
• D. $378$

Answer 1

Answer: (A)

$27$

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right].....\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]$

$\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right].......\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& \frac{n(n+1)}{2}\\ 0& 1\end{array}\right]$

Since, sum of first n natural numbers

$=\frac{n(n+1)}{2}$

$\left[\begin{array}{cc}1& \frac{n(n+1)}{2}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$

$\frac{n(n+1)}{2}=378$

$n^2+n-756=0$

$n^2+28n-27n-756=0$

$n(n+28)-27(n+28)=0$

$(n-27)(n+28)=0$

$\therefore n=27$