Mathematics — CBSE Class 10 board question

Let the first term be $a$ and common difference be $d$.

Step 1: Use given ratio of 11th and 17th terms

$$\frac{a_{11}}{a_{17}} = \frac{3}{4}$$

$$\frac{a + 10d}{a + 16d} = \frac{3}{4}$$

$$4(a + 10d) = 3(a + 16d)$$

$$4a + 40d = 3a + 48d$$

$$a = 8d$$

Step 2: Ratio of 5th term to 21st term

$$a_5 = a + 4d = 8d + 4d = 12d$$

$$a_{21} = a + 20d = 8d + 20d = 28d$$

$$\frac{a_5}{a_{21}} = \frac{12d}{28d} = \boxed{\frac{3}{7}}$$

Step 3: Ratio of sum of first 5 terms to sum of first 21 terms

$$S_5 = \frac{5}{2}(a + a_5) = \frac{5}{2}(8d + 12d) = \frac{5}{2} \times 20d = 50d$$

$$S_{21} = \frac{21}{2}(a + a_{21}) = \frac{21}{2}(8d + 28d) = \frac{21}{2} \times 36d = 378d$$

$$\frac{S_5}{S_{21}} = \frac{50d}{378d} = \boxed{\frac{25}{189}}$$

Source: Chapter 5 (Arithmetic Progressions), Section 5.3–5.4

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• Examiners award marks at each step: setting up the equation (1M), solving for $a$ in terms of $d$ (1M), finding term ratio (1M), setting up sum formula (1M), and final sum ratio (1M).
• Use $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term — it's faster than the $2a+(n-1)d$ form here.
• Always simplify the final ratio fully; leaving it unsimplified loses the accuracy mark.
• The key insight is expressing $a = 8d$ so every term cancels to a pure number ratio.