# The ratio of the sums of m and n terms of an A.P. is

Question:

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

Solution:

Let a and b be the first term and the common difference of the A.P. respectively.

According to the given condition,

$\frac{\text { Sum of } \mathrm{m} \text { terms }}{\text { Sum of } \mathrm{n} \text { terms }}=\frac{\mathrm{m}^{2}}{\mathrm{n}^{2}}$

$\Rightarrow \frac{\frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]}{\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]}=\frac{\mathrm{m}^{2}}{\mathrm{n}^{2}}$

$\Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}$ $\ldots(1)$

Putting $m=2 m-1$ and $n=2 n-1$ in (1), we obtain

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

$\Rightarrow \frac{\mathrm{a}+(\mathrm{m}-1) \mathrm{d}}{\mathrm{a}+(\mathrm{n}-1) \mathrm{d}}=\frac{2 \mathrm{~m}-1}{2 \mathrm{n}-1}$  $\ldots(2)$

$\frac{\mathrm{m}^{\text {th }} \text { term of A.P. }}{\mathrm{n}^{\text {th }} \text { term of A.P. }}=\frac{\mathrm{a}+(\mathrm{m}-1) \mathrm{d}}{\mathrm{a}+(\mathrm{n}-1) \mathrm{d}}$ $\ldots(3)$

From (2) and (3), we obtain

$\frac{m^{\text {th }} \text { term of A.P }}{n^{\text {th }} \text { term of A.P }}=\frac{2 m-1}{2 n-1}$

Thus, the given result is proved.

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