Question

The sum of the first five terms of the series \[3 + 4\frac{1}{2} + 6\frac{3}{4} + ......\]will be
A. $39\frac{9}{16}$
B. $18\frac{3}{16}$
C. $39\frac{7}{16}$
D. $13\frac{9}{16}$

Answer 1

Hint: In this question, we are to find the sum of the terms of the given series. If we find the type of the series i.e., the given series is arithmetic or geometric, we are able to find the sum of the series using the appropriate formula.

Formula Used: If the series is an Arithmetic series, then the sum of the terms in the arithmetic series is calculated by
${{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right]$ where $d={{a}_{n}}-{{a}_{n-1}}$
If the series is a geometric series, then the sum of the terms is calculated by
 ${{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ is the sum of the n terms of the series; $n$ is the number of terms; $a$ is the first term in the series; $d$ is a common difference, and $r$ is the common ratio of the terms in the series.
Complete Step-by-step Solution:
The given series is \[3 + 4\frac{1}{2} + 6\frac{3}{4} + ......\]
The series contains mixed fractions. So, simplifying them and writing in the improper fractions form as
\[3+\frac{9}{2}+\frac{27}{4}+......\]
On simplifying the series,
$3+\frac{{{3}^{2}}}{2}+\frac{{{3}^{3}}}{{{2}^{2}}}+......$
Since the next terms in the sequence are multiplied by the previous terms, this is a geometric series.
Here the common ratio is
$r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align} & \Rightarrow r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
 & \text{ =}\frac{\frac{27}{4}}{\frac{9}{2}} \\
 & \text{ =}\frac{27}{4}\times \frac{2}{9} \\
 & \text{ =}\frac{3}{2} \\
\end{align}\]
Thus, the sum of the first five terms of the given geometric sequence is calculated by the formula,
${{S}_{n}}=\frac{a({{r}^{n}}-1)}{r-1}$
On substituting $a=3;n=5;r=\frac{3}{2}$, we get
$\begin{align}
  & {{S}_{5}}=\frac{3\left( {{\left( \frac{3}{2} \right)}^{5}}-1 \right)}{\frac{3}{2}-1} \\
 & \text{ }=\frac{3\left( \frac{243}{32}-1 \right)}{\frac{1}{2}} \\
 & \text{ }=3\left( \frac{211}{32} \right)\times 2 \\
 & \text{ }=\frac{633}{16} \\
\end{align}$
The obtained sum is the improper fraction. So, writing in the mixed fraction form as ${{S}_{5}}=39\frac{9}{16}$.
Thus, Option (A) is correct.

Note: Here the given series is a geometric series since the second term is multiplied by the first term, the third term is multiplied by the second term, and so on. So, we can find the common ratio from this easily and use it to find the sum of the required number of terms.