Question

the ratio of the sums of m terms and n terms of an ap is m² is to n² show the ratio of the mth term and nth term is 2m-1:2n-1

Answer 1

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Sum of m terms of an A.P. = m/2 [2a + (m -1)d]
Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

m/2 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m:n

⇒ [2a + md - d] / [2a + nd - d] = m/n
⇒ 2an + mnd - nd = 2am + mnd - md
⇒ 2an - 2am = nd - md
⇒ 2a (n -m) = d(n - m)
⇒ 2a = d

Ratio of m th term to nth term:
[a + (m - 1)d] / [a + (n - 1)d]
= [a + (m - 1)2a] / [a + (n - 1)2a]
= a [1 + 2m - 2] / a[1 + 2n -2]
= (2m - 1) / (2n -1)

So, the ratio of mth term and the nth term of the arithmetic series is (2m - 1):(2n -1).

 

Answer 2

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).

we know that:-

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

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

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