Find the value of $$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{47.50}$$ .​

Correct option is C

Solution:
S =$$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \cdots + \frac{1}{47 \cdot 50}$$​
Concept Used:
Each term in the series is of the form:
​$$\frac{1}{a_n \cdot a_{n+1}}$$​
​where the sequence $$a_n$$​ is defined by:
​$$a_1 = 1, \quad a_{n+1} = a_n + 3$$​
This is an arithmetic sequence with first term 1 and common difference 3.

Solution:
​$$\frac{1}{a_n \cdot a_{n+1}} = \frac{1}{3} \left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right)$$​​

Substituting $$a_n = 1 + 3(n-1):$$​
​$$\frac{1}{a_n \cdot a_{n+1}} = \frac{1}{3} \left( \frac{1}{1 + 3(n-1)} - \frac{1}{1 + 3n} \right)$$​​

S = $$\frac{1}{3} \left[ \left( \frac{1}{1} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{10} \right) + \cdots + \left( \frac{1}{47} - \frac{1}{50} \right) \right]$$​

S $$= \frac{1}{3} \left( 1 - \frac{1}{50} \right) = \frac{1}{3} \cdot \frac{49}{50} = \frac{49}{150}$$​