A) 2
B) 1
C) 3
D) 4
Correct Answer: A
Solution :
\[\therefore {{n}^{th}}\]term of series \[=a{{r}^{n-1}}\]\[=a\,{{(3)}^{n-1}}=486\] ?..(i) and sum of n terms of series. \[{{S}_{n}}=\frac{a({{3}^{n}}-1)}{3-1}\]\[=728\,(\because \,r>1)\] ?..(ii) From (i), \[a\left( \frac{{{3}^{n}}}{3} \right)=486\] or \[a{{.3}^{n}}=3\times 486=1458\] From (ii), \[a{{.3}^{n}}-a=728\times 2\] or \[a{{.3}^{n}}-a=1456\] \[1458-a=1456\] Þ \[a=2\].
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