Mathematics — CBSE Class 10 board question

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

Given: $\dfrac{a_{11}}{a_{18}} = \dfrac{2}{3}$

$$\frac{a + 10d}{a + 17d} = \frac{2}{3}$$

$$3(a + 10d) = 2(a + 17d)$$

$$3a + 30d = 2a + 34d$$

$$a = 4d$$

Ratio of 5th term to 21st term:

$$\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3}$$

$$\boxed{a_5 : a_{21} = 1 : 3}$$

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

$$S_5 = \frac{5}{2}[2a + 4d] = \frac{5}{2}[8d + 4d] = \frac{5}{2} \times 12d = 30d$$

$$S_{21} = \frac{21}{2}[2a + 20d] = \frac{21}{2}[8d + 20d] = \frac{21}{2} \times 28d = 294d$$

$$\frac{S_5}{S_{21}} = \frac{30d}{294d} = \frac{5}{49}$$

$$\boxed{S_5 : S_{21} = 5 : 49}$$

Source: Chapter 5, Section 5.3 (nth term and Sum of AP)

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• Examiners award marks for: setting up the ratio equation correctly (1 mark), solving to get $a = 4d$ (1 mark), finding $a_5 : a_{21}$ (1 mark), and finding $S_5 : S_{21}$ (2 marks).
• Key formulas: $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$.
• Substitute $a = 4d$ throughout — this is the critical step. Show substitution clearly for full marks.