The ratio of the 10th term to its 30th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of A.P.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Given: $\dfrac{a_{10}}{a_{30}} = \dfrac{1}{3}$ and $S_6 = 42$
Setting up equations:
$$a_{10} = a + 9d, \quad a_{30} = a + 29d$$
$$\frac{a + 9d}{a + 29d} = \frac{1}{3}$$
$$3a + 27d = a + 29d$$
$$2a = 2d \implies a = d \quad \text{...(1)}$$
For $S_6 = 42$:
$$\frac{6}{2}[2a + 5d] = 42$$
$$3[2a + 5d] = 42 \implies 2a + 5d = 14 \quad \text{...(2)}$$
Substituting (1) into (2):
$$2d + 5d = 14 \implies 7d = 14 \implies d = 2$$
$$a = 2$$
First term $a = 2$, common difference $d = 2$.
Source: Arithmetic Progressions, Chapter 5
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Explanation
- Use $a_n = a + (n-1)d$ to form the ratio equation, cross-multiply, and simplify to get the first relation between $a$ and $d$.
- Use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ for the second equation.
- Solve the two simultaneous equations. Here both give $a = d = 2$, so verify by substitution.
- Show all steps clearly — each equation and substitution carries marks.