Question

If the ratio of sum of the first m and n terms of an AP is m² : n², show that the ratio of its  m^th and n^th terms is (2m − 1) : (2n − 1) ?

Answer 1

Let the first term and the common difference of the AP be a and d, respectively.
Therefore,
Sum of the first m terms of the AP,

${\displaystyle {\text{S}}_{\text{m}}=\frac{\text{m}}{2}[2\text{a}+\left(\text{m - 1}\right)\text{d}]}$

Sum of the first n terms of the AP,

${\displaystyle {\text{S}}_{\text{n}}=\frac{\text{n}}{2}[2\text{a}+\left(\text{n - 1}\right)\text{d}]}$

It is given that

${\displaystyle \frac{{\text{S}}_{\text{m}}}{{\text{S}}_{\text{n}}}=\frac{\frac{\text{m}}{2}[2\text{a}+\left(\text{m - 1}\right)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+\left(\text{n - 1}\right)\text{d}]}}$

${\displaystyle \Rightarrow \frac{[2\text{a}+\left(\text{m - 1}\right)\text{d}]}{[2\text{a}+\left(\text{n - 1}\right)\text{d}]}=\frac{\text{m}}{\text{n}}}$

⇒ 2an + mnd - nd = 2am + nmd - md

⇒ 2an - 2am = nd - md

⇒ 2a(n - m) = d(n - m)

⇒ 2a = d

Now,

${\displaystyle \frac{{\text{T}}_{\text{m}}}{{\text{T}}_{\text{n}}}=\frac{\text{a}+(\text{m}-1)\text{d}}{\text{a}+(\text{n}-1)\text{d}}}$

${\displaystyle \Rightarrow \frac{{\text{T}}_{\text{m}}}{{\text{T}}_{\text{n}}}=\frac{\text{a}+\left(\text{m - 1}\right)\times 2\text{a}}{\text{a}+(\text{n}-1)\times 2\text{a}}}$

${\displaystyle \Rightarrow \frac{{\text{T}}_{\text{m}}}{{\text{T}}_{\text{n}}}=\frac{2\text{m}-1}{\text{2n}-1}}$

Hence, the ratio of the mth term to the nth term is (2m − 1) : (2n − 1).