# The ratio of the

As we know that sum of n terms of an A.P is:

Sn= × (2a + (n – 1)d)

So, sum of m terms of an A.P is:

Sm = × (2a + (m – 1)d)

From the question we know that, =  =  =  = m × (2a + (n – 1)d) = n × (2a + (m – 1)d)

2ma + m(n – 1)d = 2na + n(m – 1)d

2ma + md(n – 1) = 2na + nd(m – 1)

2ma + mnd – md = 2na + mnd – nd

2ma – md = 2na – nd

2ma – 2na = md – nd

2a(m – n) = d(m – n)

2a = d

Now, mth term of the given A.P is Tm = a + (m – 1)d

Now, nth term of the given A.P is Tn = a + (n – 1)d

Now ratio between these 2 terms is: = Now we have, 2a = d

So we get, =  =  =  =  = the ratio of mth term of the given A.P to its nth Term is:

Tm : Tn = 2m – 1 : 2n – 1

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