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

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Step 1: $a = 45$, $d = -6$, $S_n = 180$
Step 2: $180 = \frac{n}{2}[2 	imes 45 + (n - 1)(-6)]$
Step 3: $\Rightarrow 360 = 96n - 6n^2$
Step 4: $\Rightarrow 6n^2 - 96n + 360 = 0$
Step 5: $\Rightarrow n^2 - 16n + 60 = 0 \Rightarrow (n - 10)(n - 6) = 0 \Rightarrow n = 10, 6$
Step 6: We get two values of $n$ as sum of 7th term to 10th terms is zero as some terms are negative and some are positive.

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