Mathematics — CBSE Class 10 board question

Given: AP = 45, 39, 33, …
Here, $a = 45$, $d = 39 - 45 = -6$, $S_n = 180$

Using the sum formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

$$180 = \frac{n}{2}[2(45) + (n-1)(-6)]$$

$$360 = n[90 - 6n + 6]$$

$$360 = n[96 - 6n]$$

$$360 = 96n - 6n^2$$

$$6n^2 - 96n + 360 = 0$$

$$n^2 - 16n + 60 = 0$$

$$(n - 6)(n - 10) = 0$$

$$\therefore n = 6 \quad \text{or} \quad n = 10$$

Explanation of double answer:

Both answers are valid. When $n = 6$, $S_6 = 180$.
When $n = 10$, $S_{10} = 180$ also.

This happens because the AP has a positive first term but a negative common difference ($d = -6$), so the terms decrease and eventually become negative. The terms from 7th to 10th are negative (e.g., $a_7 = 45 + 6(-6) = 9$; $a_8 = 3$; $a_9 = -3$; $a_{10} = -9$). The negative terms from 7th onward cancel the extra sum added after the 6th term, making $S_{10} = S_6 = 180$.

Source: Chapter 5, Section 5.3 — Sum of First n Terms of an AP

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• Key formula: $S_n = \frac{n}{2}[2a + (n-1)d]$; substituting correctly gives a quadratic in $n$.
• Both roots are positive integers, so both are accepted — write both.
• Examiner expects the double-answer explanation: The terms after $n = 6$ are small/negative (AP is decreasing), so their net contribution is zero over terms 7–10, giving the same sum again. Always verify both values of $n$ are natural numbers.
• Award of marks: ~2 marks for setting up and solving the quadratic, 1 mark each for the two values of $n$, 1 mark for the explanation.