Question

$6$th term of the sequence $\frac{7}{3},\frac{35}{6},\frac{121}{12},\frac{335}{24}.....$ is

• A. $\frac{2113}{96}$
• B. $\frac{2112}{96}$
• C. $\frac{2111}{96}$
• D. $\frac{865}{48}$

Answer 1

The correct option is C

$\frac{2111}{96}$

Explanation for the correct option:

Step 1: Expand the series:

Break each term into two parts.

$2+\frac{1}{3},6-\frac{1}{6},10+\frac{1}{12}....,$

Observe that the first part of each term is in AP with the common difference $=4$

Also, the second part of each term is in GP with the common ratio $=-\frac{1}{2}$

Step 2: Find the $6$th term of the sequence.

Estimate the term by finding $6$th term of both parts of the term.

$$\begin{gathered} T_6=2+(6-1)4+\frac{1}{3}·{\left(-\frac{1}{2}\right)}^{6-1}\left[{\mathbf{\text{∵T}}}_{\mathbf{n}}\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{d}\mathbf{}\mathbf{for}\mathbf{}\mathbf{an}\mathbf{}\mathbf{AP}\mathbf{,}\mathbf{}{\mathbf{ar}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{for}\mathbf{}\mathbf{an}\mathbf{}\mathbf{GP}\right] \\ =2+20-\frac{1}{96} \\ =\frac{2111}{96} \end{gathered}$$

Hence, option (C) is the correct answer.