Question

The ratio of the sums of first m & find n terms of an Ap is m²: n².show that the ratio of it's mth &n th term is(2m-1):(2n-1).


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Answer 1

Given

• The ratio of the sums of first m & n terms of an Ap is m²: n²

To Prove

• the ratio of it's mth &n th term is(2m-1):(2n-1).

Concept used

• $\rm \: s_{n} = \dfrac{n}{2} (2a + (n - 1)d)$
• $\rm \: a_{n} = a + (n - 1)d$

Proof

We have given the ratio

Therefore,

$\rm \: \dfrac{ s_{m} }{ s_{n}} = \dfrac{ {m}^{2} }{ {n}^{2} } \\ \\ \implies \rm \: \frac{ \dfrac{m}{2}(2a + (m - 1)d) }{ \dfrac{n}{2}(2a + (n - 1)d) } = \frac{ {m}^{2} }{ {n}^{2} } \\ \\ \implies \rm \: \frac{m(2a + (m - 1)d)}{n(2a + (n - 1)d)} = \frac{ {m}^{2} }{ {n}^{2} } \\ \\ \implies \rm \: \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{ {m}^{2} }{ {n}^{2} } \times \frac{n}{m}$

$\rm\implies \: n(2a + (m - 1)d )= \: m(2a + (n - 1)d) \\ \\ \rm \implies \: 2an + (m - 1)dn = 2am + (n - 1)dm \\ \\ \rm \implies \: 2an - 2am = dmn - dmn - dm + dn \\ \\ \implies \rm \: 2a(n - m) = d(n - m) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \boxed{ \bf \: 2a = d} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now we find the ratio of mth and nth term

therefore

$\rm \dfrac{ a_{m}}{ a_{n} } = \dfrac{a + (m - 1)d}{a + (n - 1)d} \\ \\ \implies \rm \: \frac{a_{m}}{a_{n}} = \frac{a + 2am - 2a}{a + 2an - 2a} \\ \\ \implies \rm \: \frac{a_{m}}{a_{n}} = \frac{2am - a}{2an - a} \: \: \: \: \: \: \\ \\ \implies \rm \: \frac{a_{m}}{a_{n}} = \frac{a(2m - 1)}{a(2n - 1)} \\ \\ \implies \boxed{ \bf \: \frac{a_{m}}{a_{n}} = \frac{2m - 1}{2n - 1} }$

Thus,the ratio is 2m-1:2n-1

Hence proved