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 \[14m-6:8m+23\]
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