A) \[\frac{1}{5}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{9}\]
Correct Answer: B
Solution :
First term of an A.P. = 1, let Common difference = d \[\therefore {{T}_{2}}=a+d,\,\,{{T}_{10}}=a+9d,{{T}_{34}}=a+33d\] \[\therefore {{(a+9d)}^{2}}=(a+d)(a+33d)\] Þ \[{{a}^{2}}+81{{d}^{2}}+18ad={{a}^{2}}+ad+33ad+33{{d}^{2}}\] Put \[a=1\] \[\Rightarrow 1+81{{d}^{2}}+18d=1+d+33d+33{{d}^{2}}\] Þ \[48{{d}^{2}}-16d=0\] \[\Rightarrow 16d(3d-1)=0\] Þ \[d=0,\,\,d=1/3\].
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