Question

If $\frac{1}{2.7} + \frac{1}{7.12} + \frac{1}{12.17} + \dots$ to 10 terms = k, then k =

• A. $\frac{2}{51}$
• B. $\frac{5}{51}$
• C. $\frac{5}{52}$
• D. $\frac{1}{26}$

Hint:

For a series with terms of the form $\frac{1}{a_n a_{n+1}}$ where $a_n$ is an AP, you can use the shortcut: $T_n = \frac{1}{d} \left(\frac{1}{a_n} - \frac{1}{a_{n+1}}\right)$, where $d$ is the common difference. The sum of $N$ terms is then $S_N = \frac{1}{d} \left(\frac{1}{a_1} - \frac{1}{a_{N+1}}\right)$. Here, the AP is $2, 7, 12, ...$ with $d=5$. The sum is $\frac{1}{5} (\frac{1}{\text{first term}} - \frac{1}{\text{last term}})$. The first term is 2. The last term in the sum is the second part of $T_{10}$, which is $5(10)+2=52$. So $S_{10} = \frac{1}{5}(\frac{1}{2} - \frac{1}{52})$.

Answer 1

The Correct Option is C

Step 1: Identify the General Term: The denominators are products of two numbers. Let's look at the pattern: First factors: 2, 7, 12, 17, ... (Arithmetic Progression with \( a=2, d=5 \)). \( n \)-th term: \( 2 + (n-1)5 = 5n - 3 \). Second factors: 7, 12, 17, 22, ... (Arithmetic Progression with \( a=7, d=5 \)). \( n \)-th term: \( 7 + (n-1)5 = 5n + 2 \). So, the series is \( \sum_{n=1}^{10} \frac{1}{(5n-3)(5n+2)} \).
Step 2: Telescoping Series Method: We can decompose the general term using partial fractions: \[ \frac{1}{(5n-3)(5n+2)} = \frac{1}{5} \left( \frac{1}{5n-3} - \frac{1}{5n+2} \right) \] Check: \( \frac{1}{5} \frac{(5n+2)-(5n-3)}{(5n-3)(5n+2)} = \frac{1}{5} \frac{5}{\text{denom}} = \frac{1}{\text{denom}} \). Correct.
Step 3: Summing the Series: \( S_{10} = \frac{1}{5} \sum_{n=1}^{10} \left( \frac{1}{5n-3} - \frac{1}{5n+2} \right) \) Expanding the terms: \( n=1: \frac{1}{2} - \frac{1}{7} \) \( n=2: \frac{1}{7} - \frac{1}{12} \) ... \( n=10: \frac{1}{5(10)-3} - \frac{1}{5(10)+2} = \frac{1}{47} - \frac{1}{52} \) All intermediate terms cancel out (Telescoping Sum). \( S_{10} = \frac{1}{5} \left( \frac{1}{2} - \frac{1}{52} \right) \)
Step 4: Final Calculation: \[ S_{10} = \frac{1}{5} \left( \frac{26 - 1}{52} \right) = \frac{1}{5} \left( \frac{25}{52} \right) \] \[ k = \frac{5}{52} \]