Sequence And Series Question 223
Question: The sum of the first five terms of the series $ 3+4\frac{1}{2}+6\frac{3}{4}+…… $ will be
Options:
A) $ 39\frac{9}{16} $
B) $ 18\frac{3}{16} $
C) $ 39\frac{7}{16} $
D) $ 13\frac{9}{16} $
Show Answer
Answer:
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[ {{( \frac{3}{2} )}^{5}}-1 ]}{\frac{3}{2}-1}=\frac{1[ \frac{3^{5}}{32}-1 ]}{\frac{1}{2}} $ $ =6[ \frac{243-32}{32} ]=\frac{211\times 3}{16}=\frac{633}{16}=39\frac{9}{16} $ .
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