The ratio of the sums of first $m$ and first $n$ terms of an AP is $m^2 : n^2$. Show that the ratio of its $m^{\text{th}}$ and $n^{\text{th}}$ terms is $(2m - 1) : (2n - 1)$.
Step-by-Step Solution
Key Concept: From $\dfrac{S_m}{S_n} = \dfrac{m^2}{n^2}$, simplify to show $a = d/2 \Rightarrow d = 2a$, then compute $\dfrac{a_m}{a_n} = \dfrac{a + (m-1)d}{a + (n-1)d}$.
Given $\dfrac{S_m}{S_n} = \dfrac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \dfrac{m^2}{n^2}$. [1.0 Mark]
Cancel $m/n$: $\dfrac{2a + (m-1)d}{2a + (n-1)d} = \dfrac{m}{n} \Rightarrow n[2a + (m-1)d] = m[2a + (n-1)d]$. [1.0 Mark]
$2an + nmd - nd = 2am + mnd - md \Rightarrow 2a(n - m) = d(n - m) \Rightarrow d = 2a$. [1.5 Marks]
Now ratio of $m^{\text{th}}$ and $n^{\text{th}}$ terms:
$\dfrac{a_m}{a_n} = \dfrac{a + (m-1)d}{a + (n-1)d} = \dfrac{a + (m-1)(2a)}{a + (n-1)(2a)} = \dfrac{a(1 + 2m - 2)}{a(1 + 2n - 2)} = \dfrac{2m - 1}{2n - 1}$. Proved! [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Setting up sum ratio equation: 1.0 Mark
Simplifying equation: 1.0 Mark
Proving $d = 2a$: 1.5 Marks
Evaluating term ratio $a_m/a_n = (2m-1)/(2n-1)$: 1.5 Marks
Correct Answer: