The value of s = 
I started this question by making an A.P as the common difference is same and got the answer that I need number of terms to proceed further but my valie for number of terms is coming in fraction that is not possible i tried many times but end up with the same value so I can't proceed further as I don't know the value of n to put in the sum of Arithmetic Progression series?
Answer
Hint:
The general term is $$\frac{1}{(5r-3)(5r+2)}$$
We use partial fraction decomposition, so:
$$\frac{1}{(5r-3)(5r+2)}=\frac{A}{5r-3}+\frac{B}{5r+2}$$
Can you find $A$ and $B$? (Hint 2: $A+B=0$)
Because $A+B=0$, can you see that between the decomposition of $$\frac{1}{(5r-3)(5r+2)}$$ and $$\frac{1}{(5(r+1)-3)(5(r+1)+2)}=\frac{1}{(5r+2)(5r+7)}$$
some terms are cancelled? Figure out which ones will not be cancelled and sum them.
This method is known as telescoping
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