Mathematics — Standard (041) — CBSE Class 10 board question

Given AP: 65, 61, 57, 53, …

Here, $a = 65$, $d = 61 - 65 = -4$

Using the formula: $a_n = a + (n-1)d$

For the first negative term, we need $a_n < 0$:

$$65 + (n-1)(-4) < 0$$

$$65 - 4n + 4 < 0$$

$$69 - 4n < 0$$

$$4n > 69$$

$$n > 17.25$$

Since $n$ must be a whole number, the smallest value is $n = 18$.

The 18th term is the first negative term.

Verification: $a_{18} = 65 + 17 \times (-4) = 65 - 68 = -3$ ✓

Source: Chapter 5, Section 5.3 (nth term formula)

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• Set up $a_n < 0$ using $a_n = a + (n-1)d$ and solve the inequality for $n$.
• Since $n$ must be a natural number, always round up to the next whole number (here 17.25 → 18).
• Verify by substituting $n = 18$ to confirm the term is negative and check $n = 17$ gives a positive value ($a_{17} = 65 - 64 = 1 > 0$).
• Examiners expect the inequality approach, not trial-and-error.