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Sum

The ratio of the sums of m and n terms of an A.P. is m^{2} : n^{2}. Show that the ratio of the m^{th} and n^{th} terms is (2m – 1) : (2n – 1)

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#### Solution

Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by

`S_m = \frac { m }{ 2 } [2a + (m – 1) d], and S_n = \frac { n }{ 2 }[2a + (n – 1) d]`

respectively. Then,

`S_m/S_n=m^2/n^2=>(m/2[2a+(m-1)d])/(n/2[2a+(n-1)d])=m^2/n^2`

`=>(2a+(m-1)d)/(2a+(n-1)d)=m/n`

⇒ [2a + (m – 1) d] n = {2a + (n – 1) d} m

⇒ 2a (n – m) = d {(n – 1) m – (m – 1) n}

⇒ 2a (n – m) = d (n – m)

⇒ d = 2a

`T_m/T_n=(a+(m-1)d)/(a+(n-1)d)`

`=\frac{a+(m-1)2a}{a+(n-1)2a}=\frac{2m-1}{2n-1}`

Thus, the ratio of its mth and nth terms is 2m – 1 : 2n – 1.

Concept: Sum of First n Terms of an A.P.

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