A) 53 : 155
B) 27 : 77
C) 29 : 83
D) 31 : 89
Correct Answer: A
Solution :
: We have, \[\frac{{{S}_{n}}}{S{{'}_{n}}}=\frac{2n+3}{6n+5},\]where\[{{S}_{n}}\]and \[S_{n}^{'}\]are the sum of n terms of given arithmetic series. \[\Rightarrow \]\[\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d']}=\frac{2n+3}{6n+5}\] \[\Rightarrow \frac{a+\left( \frac{n-1}{2} \right)d}{a'+\left( \frac{n-1}{2} \right)d'}=\frac{2n+3}{6n+5}\] Now, put \[\frac{n-1}{2}=12\]or n = 25 then\[\frac{a+12d}{a'+12d'}=\frac{2(25)+3}{6(25)+5}\]\[\Rightarrow \]\[\frac{{{T}_{13}}}{T{{'}_{12}}}=\frac{53}{155}\]
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