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Sum of n Terms

The ratio of ...

Question

The ratio of the sums of m and n terms of an A.P. is $m^{2} : n^{2}$ , then the ratio of the $m^{t h}$ and $n^{t h}$ terms is

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Solution

Sum of $m$ terms of an $A . P . = \frac{m}{2} [2 a + (m - 1) d]$
Sum of $n$ terms of an $A . P . = \frac{n}{2} [2 a + (n - 1) d]$

$\frac{\frac{m}{2} [2 a + (m - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = m : n$

$\Rightarrow \frac{[2 a + m d - d]}{[2 a + n d - d]} = \frac{m}{n}$

$\Rightarrow 2 a n + m n d - n d = 2 a m + m n d - m d$
$\Rightarrow 2 a n - 2 a m = n d - m d$
$\Rightarrow 2 a (n - m) = d (n - m)$
$\Rightarrow 2 a = d$

Ratio of $m^{t h}$ term to $n^{t h}$ term:
$\frac{[a + (m - 1) d]}{[a + (n - 1) d]}$

$= \frac{[a + (m - 1) 2 a]}{[a + (n - 1) 2 a]}$

$= \frac{a [1 + 2 m - 2]}{a [1 + 2 n - 2]}$

$= \frac{(2 m - 1)}{(2 n - 1)}$

So, the ratio of $m^{t h}$ term and the $n^{t h}$ term of the arithmetic series is
$= (2 m - 1) : (2 n - 1)$

Hence, this is the answer.

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