For a sequence \[,\ {{a}_{1}}=2\] and \[\frac{{{a}_{n+1}}}{{{a}_{n}}}=\frac{1}{3}\]. Then \[\sum\limits_{r=1}^{20}{{{a}_{r}}}\] is

\[3\left( 1-\frac{1}{{{3}^{20}}} \right)\]

For a sequence \[,\ {{a}_{1}}=2\] and \[\frac{{{a}_{n+1}}}{{{a}_{n}}}=

1.	For a sequence \[,\ {{a}_{1}}=2\] and \[\frac{{{a}_{n+1}}}{{{a}_{n}}}=\frac{1}{3}\]. Then \[\sum\limits_{r=1}^{20}{{{a}_{r}}}\] is
A.	\[\frac{20}{2}[4+19\times 3]\]
B.	\[3\left( 1-\frac{1}{{{3}^{20}}} \right)\]
C.	\[2(1-{{3}^{20}})\]
D.	None of these
Answer» C. \[2(1-{{3}^{20}})\]

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For a sequence \[,\ {{a}_{1}}=2\] and \[\frac{{{a}_{n+1}}}{{{a}_{n}}}=\frac{1}{3}\]. Then \[\sum\limits_{r=1}^{20}{{{a}_{r}}}\] is

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