question_answer

The ratio of the sums of first m and first n terms of an A. P. is \[{{m}^{2}}:{{n}^{2}}\].

Show that the ratio of its mth and nth terms is \[2(m-1):(2n-1)\].

Answer:

Let a be first term and d is common difference.

Then,                \[\frac{{{S}_{m}}}{{{S}_{n}}}=\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \]   \[n[2a+(m-1)d]=m[2a+(n-1)d]\]

\[\Rightarrow \]   \[2an+nd(m-1)=2am+md(n-1)\]

\[\Rightarrow \]            \[2a(n-m)=[m(n-1)-n(m-1)d\]

\[=(mn-m-mn+n)d\]

\[2a(n-n)=(n-m)d\]

\[2a=d\]

Now,                     \[\frac{{{T}_{m}}}{{{T}_{n}}}=\frac{a+(m-1)d}{a+(n-1)d}\]

\[=\frac{a+(m-1)2a}{a+(n-1)2a}=\frac{2m-1}{2n-1}\]                     Hence Proved.