A) \[39\frac{9}{16}\]
B) \[18\frac{3}{16}\]
C) \[39\frac{7}{16}\]
D) \[13\frac{9}{16}\]
Correct Answer: A
Solution :
Given series is \[3+4\frac{1}{2}+6\frac{3}{4}+........=3+\frac{9}{2}+\frac{27}{4}+.....\] \[=3+\frac{{{3}^{2}}}{2}+\frac{{{3}^{3}}}{4}+\frac{{{3}^{4}}}{8}+\frac{{{3}^{5}}}{16}+.....\](in G.P.) Here\[a=3,\ r=\frac{3}{2}\] , then sum of the five terms \[{{S}_{5}}=\frac{a({{r}^{n}}-1)}{r-1}=\frac{3\left[ {{\left( \frac{3}{2} \right)}^{5}}-1 \right]}{\frac{3}{2}-1}=\frac{1\left[ \frac{{{3}^{5}}}{32}-1 \right]}{\frac{1}{2}}\] \[=6\left[ \frac{243-32}{32} \right]=\frac{211\times 3}{16}=\frac{633}{16}=39\frac{9}{16}\].
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