A) 10
B) 20
C) 30
D) 40
E) 50
Correct Answer: A
Solution :
Let \[{{S}_{1}}=a,{{S}_{2}}=a+d,{{S}_{3}}=a+2d....\] \[{{S}_{101}}=a+100d\]are in AP. Given, \[\frac{1}{{{S}_{1}}{{S}_{2}}}+\frac{1}{{{S}_{2}}{{S}_{3}}}+.....+\frac{1}{{{S}_{100}}{{S}_{101}}}=\frac{1}{6}\] \[\Rightarrow \] \[\frac{1}{a(a+d)}+\frac{1}{(a+d)(a+2d)}\] \[+....+\frac{1}{(a+99d)(a+100d)}=\frac{1}{6}\] \[\Rightarrow \] \[\frac{1}{d}\left\{ \frac{1}{a}-\frac{1}{a+d} \right\}+\frac{1}{d}\left\{ \frac{1}{a+d}-\frac{1}{a+2d} \right\}\] \[+...+\frac{1}{d}\left\{ \frac{1}{a+99d}-\frac{1}{a+100d} \right\}=\frac{1}{6}\] \[\Rightarrow \] \[\frac{1}{a}-\frac{1}{a+100d}=\frac{d}{6}\] \[\Rightarrow \] \[\frac{100d}{a(a+100d)}=\frac{d}{6}\] \[\Rightarrow \] \[600=a(a+100d)\] \[\Rightarrow \] \[{{S}_{101}}=\frac{600}{a}\] ?.. (i) Given that, \[{{S}_{1}}+{{S}_{101}}=50\] \[a+\frac{600}{a}=50\] \[\Rightarrow \] \[{{a}^{2}}-50a+600=0\] \[\Rightarrow \] \[{{a}^{2}}-30a-20a+600=0\] \[\Rightarrow \] \[a(a-30)-20(a-30)=0\] \[\Rightarrow \] \[(a-30)(a-20)=0\] \[\Rightarrow \] \[a=20,30\] So, \[{{S}_{1}}=20\]or \[{{S}_{1}}=30\] and \[{{S}_{101}}=a+100d=\frac{600}{a}\] [from Eq. (i)] When \[a=20,20+100d=\frac{600}{20}=30\] \[100d=10\] \[\Rightarrow \] \[d=\frac{1}{10}\] When \[a=30,30+100d=\frac{600}{30}=20\] \[d=-\frac{1}{10}\] So, when\[{{S}_{1}}=20,\]then, \[{{S}_{101}}=a+100d\] \[=20+\frac{100}{10}=20\] when\[{{S}_{1}}=30,\]then\[{{S}_{101}}=a+100d\] \[=30-\frac{100}{10}=20\] Hence, (i) When\[{{S}_{1}}=20,{{S}_{101}}=30\] \[|{{S}_{1}}-{{S}_{101}}|=|20-30|=10\] (ii) When\[{{S}_{1}}=30,{{S}_{101}}=20\] \[|{{S}_{1}}-{{S}_{101}}|=|30-20|=10\]
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