The $8^{\text{th}}$ term of an A.P. is zero. Prove that its $38^{\text{th}}$ term is triple its $18^{\text{th}}$ term.
Step-by-Step Solution
Key Concept: $a_8 = a + 7d = 0 \Rightarrow a = -7d$. $a_{18} = a + 17d = 10d$. $a_{38} = a + 37d = 30d$. Since $30d = 3(10d)$, $a_{38} = 3 a_{18}$.
$a + 7d = 0 \Rightarrow a = -7d$. [0.5 Mark]
$a_{18} = -7d + 17d = 10d$. [0.5 Mark]
$a_{38} = -7d + 37d = 30d = 3(10d) = 3 a_{18}$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding $a = -7d$: 0.5 Mark
Evaluating $a_{18} = 10d$ and $a_{38} = 30d$: 0.5 Mark
Concluding $a_{38} = 3 a_{18}$: 1.0 Mark
Correct Answer: