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The value of s = enter image description here 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?

Ankit Kumar
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    Why would you think this was an arithmetic progression? – lulu Jun 21 '19 at 16:55
  • As the common difference is same and it is the question of chapter A.P – Ankit Kumar Jun 21 '19 at 16:57
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    The common difference is the same? Are you sure? What value are you getting for the common difference? – lulu Jun 21 '19 at 16:59
  • The terms of this series are of the form $\frac1{(5k+2)(5(k+1)+2))}$ for $k\geq0$. Decompose this fraction by partial fraction decomposition to get a telescoping series. – Rushabh Mehta Jun 21 '19 at 17:03
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    In case lulu's comments were not blatant enough, this is not an arithmetic progression. An arithmetic progression is one of the form $a+nk$ where $a$ is the first term, $n$ is the index, and $k$ is the common difference between terms. There is not a common difference here. For instance the difference between the first two is $\frac{1}{84}-\frac{1}{14}=-\frac{5}{84}$ while the difference between the second two is $\frac{1}{204}-\frac{1}{84}=-\frac{5}{714}$, a different number. There is not a common factor here either so it is also not geometric. – JMoravitz Jun 21 '19 at 17:14

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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

Rhys Hughes
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