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

The correct option is A

$39\frac{9}{16}$

Explanation for the correct answer:

The given sum can be written as

$3+4\frac{1}{2}+6\frac{3}{4}+...$$=3+\frac{9}{2}+\frac{27}{4}+...$

$=3+3\left(\frac{3}{2}\right)+3{\left(\frac{3}{2}\right)}^2+...$

Hence, the given sum is a geometric progression with first term of the G.P $a=3$ and common ratio $r=\frac{3}{2}$

The sum of $\mathrm{n}$ terms of a geometric progression can be given as

${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}$

Substitute the values of $\mathrm{a},\mathrm{r},\mathrm{n}$

$S_5=\frac{3\left[{\left({\displaystyle \frac{3}{2}}\right)}^5-1\right]}{{\displaystyle \frac{3}{2}}-1}$

$\Rightarrow$ $S_5=\frac{3\times {\displaystyle \frac{211}{32}}}{{\displaystyle \frac{1}{2}}}$

$\Rightarrow$ $S_5=\frac{633}{16}$

$\Rightarrow$ $S_5=39\frac{9}{16}$

The sum of the first five numbers of the given series is $39\frac{9}{16}$.

Hence, option A is the correct answer.