question_answer

If the ratio of the sum of first n terms of two A.P?s is \[(7n+1):(4n+27)\], find the ratio of their \[{{m}^{th}}\] terms.

Answer:

Let the sum of first n terms of two A.P?s be \[{{S}_{n}}\] and \[{{S}_{n}}'.\]

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

\[=\frac{7n+1}{4n+27}\]

\[\frac{a+\left( \frac{n-1}{2} \right)d}{a'+\left( \frac{n-1}{2} \right)d'}=\frac{7n+1}{4n+27}\]                                            ?(i)

Also, let \[{{m}^{th}}\] term of two A.P?s be \[{{T}_{m}}\] and \[{{T}_{m}}\]?

\[\frac{{{T}_{m}}}{{{T}_{m}}'}=\frac{a+(m-1)d}{a'+(m-1)d'}\]

Replacing \[\frac{n-1}{2}\] by \[m-1\] in (i), we get

\[\frac{a+(m-1)d}{a'+(m-1)d'}=\frac{7(2m-1)+1}{4(2m-1)+27}\]

\[[\because \,n-1=2(m-1)\Rightarrow n=2m-2+1=2m-1]\]

\[\therefore \frac{{{T}_{m}}}{{{T}_{m}}'}=\frac{14m-7+1}{8m-4+27}=\frac{14m-6}{8m+23}\]

\[\therefore \]  Ratio of \[{{m}^{th}}\] term of two A.P's is \[14m-6:8m+23\]