terms of two arithmatic series are in the ratio \[2n+3:6n+5\], then the ratio of their \[{{13}^{th}}\] terms is [MP PET 2004]
A) 53 : 155
B) 27 : 77
C) 29 : 83
D) 31 : 89
Correct Answer: A
Solution :
We have \[\frac{{{S}_{{{n}_{1}}}}}{{{S}_{{{n}_{2}}}}}=\frac{2n+3}{6n+5}\] Þ \[\frac{\frac{n}{2}[2{{a}_{1}}+(n-1){{d}_{1}}]}{\frac{n}{2}[2{{a}_{2}}+(n-1){{d}_{2}}]}=\frac{2n+3}{6n+5}\] Þ \[\frac{2\left[ {{a}_{1}}+\left( \frac{n-1}{2} \right)\,{{d}_{1}} \right]}{2\left[ {{a}_{2}}+\left( \frac{n-1}{2} \right)\,{{d}_{2}} \right]}=\frac{2n+3}{6n+5}\] Þ \[\frac{{{a}_{1}}+\left( \frac{n-1}{2} \right)\,{{d}_{1}}}{{{a}_{2}}+\left( \frac{n-1}{2} \right)\,{{d}_{2}}}=\frac{2n+3}{6n+5}\] Put \[n=25\] then\[\frac{{{a}_{1}}+12{{d}_{1}}}{{{a}_{2}}+12{{d}_{2}}}=\frac{2(25)+3}{6(25)+3}\]Þ \[\frac{{{T}_{{{13}_{1}}}}}{{{T}_{{{13}_{2}}}}}=\frac{53}{155}\].
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