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

Numbers divisible by 5 but not by 2 are odd multiples of 5: 5, 15, 25, 35, …, 995.

This is an AP with $a = 5$, $d = 10$, $a_n = 995$.

$$a_n = a + (n-1)d \Rightarrow 995 = 5 + (n-1) \times 10$$

$$990 = (n-1) \times 10 \Rightarrow n - 1 = 99 \Rightarrow n = 100$$

Therefore, there are 100 such natural numbers.

Source: Chapter 5, Section 5.2 (nth term of an AP)

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• Numbers divisible by 5 but not by 2 must be odd multiples of 5 (i.e., 5, 15, 25, … — skip even multiples like 10, 20, 30…).
• This forms an AP with common difference 10 (every alternate multiple of 5).
• Apply the nth term formula $a_n = a + (n-1)d$ to find $n$.
• Examiner expects you to identify the AP correctly and show the formula step clearly.