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: Understanding the Concept
The given series is a special type of series where each term can be expressed as the difference of two consecutive terms of another sequence. This is known as a telescoping series. The sum can be found by writing out the terms and observing that the intermediate terms cancel each other out.
Step 2: Key Formula or Approach
First, find the general n-th term, $T_n$, of the series. The denominators consist of products of numbers in an arithmetic progression. We can use the method of partial fractions to decompose $T_n$ into the form $\frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{b_n} \right)$.
Step 3: Detailed Explanation
The sequence of the first factors in the denominators is 2, 7, 12, 17, ... This is an Arithmetic Progression (AP) with first term $a_1=2$ and common difference $d=5$. The n-th term is $a_n = a_1 + (n-1)d = 2 + (n-1)5 = 5n-3$.
The sequence of the second factors is 7, 12, 17, 22, ... This is also an AP with first term $b_1=7$ and common difference $d=5$. The n-th term is $b_n = b_1 + (n-1)d = 7 + (n-1)5 = 5n+2$.
So, the general n-th term of the series is:
\[ T_n = \frac{1}{(5n-3)(5n+2)} \] We use partial fractions to decompose $T_n$. \[ \frac{1}{(5n-3)(5n+2)} = \frac{A}{5n-3} + \frac{B}{5n+2} \] \[ 1 = A(5n+2) + B(5n-3) \] Setting $5n-3=0 \implies n=3/5$, we get $1 = A(5(3/5)+2) \implies 1 = 5A \implies A = 1/5$. Setting $5n+2=0 \implies n=-2/5$, we get $1 = B(5(-2/5)-3) \implies 1 = -5B \implies B = -1/5$. So, the n-th term can be written as:
\[ T_n = \frac{1}{5} \left( \frac{1}{5n-3} - \frac{1}{5n+2} \right) \] We need to find the sum of the first 10 terms, $S_{10}$.
\[ S_{10} = \sum_{n=1}^{10} T_n = \frac{1}{5} \sum_{n=1}^{10} \left( \frac{1}{5n-3} - \frac{1}{5n+2} \right) \] Let's write out the terms:
For n=1: $\frac{1}{5} \left( \frac{1}{2} - \frac{1}{7} \right)$
For n=2: $\frac{1}{5} \left( \frac{1}{7} - \frac{1}{12} \right)$
For n=3: $\frac{1}{5} \left( \frac{1}{12} - \frac{1}{17} \right)$
...
For n=10: $\frac{1}{5} \left( \frac{1}{5(10)-3} - \frac{1}{5(10)+2} \right) = \frac{1}{5} \left( \frac{1}{47} - \frac{1}{52} \right)$
When we add all these terms, the intermediate terms cancel out.
\[ S_{10} = \frac{1}{5} \left[ \left(\frac{1}{2} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{12}\right) + \dots + \left(\frac{1}{47} - \frac{1}{52}\right) \right] \] \[ S_{10} = \frac{1}{5} \left( \frac{1}{2} - \frac{1}{52} \right) \] \[ S_{10} = \frac{1}{5} \left( \frac{26 - 1}{52} \right) = \frac{1}{5} \left( \frac{25}{52} \right) \] \[ S_{10} = \frac{5}{52} \] Step 4: Final Answer
The sum of the series to 10 terms is $k = \frac{5}{52}$.