A) G.P.
B) A.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
[c] \[x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}=\frac{1}{1-a};\,\,a=1-\frac{1}{x}\] \[y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}=\frac{1}{1-b};\,\,\,\,\,b=1-\frac{1}{y}\] \[z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}=\frac{1}{1-c};\,\,\,\,c=1-\frac{1}{z}\] a, b, c care in A.P. OR \[2b=a+c\] \[2\left( 1-\frac{1}{y} \right)=1-\frac{1}{x}+1-\frac{1}{y}\,\,\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\] \[\Rightarrow \,\,x,y,z\]are in H.P.You need to login to perform this action.
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