A) 2
B) 1
C) 3
D) None of these
Correct Answer: A
Solution :
[a] Let \[{{S}_{n}}\] denote the sum of n terms of an A. P. According to given \[\frac{{{S}_{n}}}{{{S}_{2n}}-{{S}_{n}}}=k\,\,\,\forall n\ge 1\Rightarrow \frac{{{S}_{1}}}{{{S}_{2}}-{{S}_{1}}}=\frac{{{S}_{2}}}{{{S}_{4}}-{{S}_{2}}}\] \[\Rightarrow \,\,\,{{S}_{1}}{{S}_{4}}-{{S}_{1}}{{S}_{2}}=S_{2}^{2}-{{S}_{1}}{{S}_{2}}\Rightarrow {{S}_{1}}{{S}_{4}}=S_{2}^{2}\] \[\Rightarrow \,\,a\frac{4}{2}[2a+(4-1)d]={{(a+a+d)}^{2}}\] \[\Rightarrow \,\,\,a(4a+6d)={{(2a+d)}^{2}}\] \[\Rightarrow \,\,\,\,2ad={{d}^{2}}\Rightarrow \,\,2a=d\] Since \[a=1,\] we get \[d=2\]
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