How many terms of the arithmetic progression 45, 39, 33, must be taken so that their sum is 180? Explain the double answer.

$\displaystyle 45$, $\displaystyle 39$, 33....... \[\begin{aligned} & a=45, d=-6 \\ & S_{n}=180 \end{aligned} \] \[180=\frac{n}{2}[2 \times 45+(n-1)(-6)] \] \(\displaystyle \Rightarrow 180=\frac{\mathrm{n}}{2}[90-6 \mathrm{n}+6]\) \[\begin{aligned} & \Rightarrow 360=96 n-6 n^{2} \\ & \Rightarrow 6 n^{2}-96 n+360=0 \\ & \Rightarrow n^{2}-16 n+60=0 \Rightarrow(n-10)(n-6)=0 \\ & n-10=0, n-6=0 \Rightarrow n=10,6 \end{aligned} \] We get two values of 'n' as sum of \(\displaystyle 7{ }^{\text {th }}\) term to \(\displaystyle 10^{\text {th }}\) term is zero as some terms are negative and some are positive.