**Q) Find the sum of π terms of the series (4 β1/π) + (4 β2/π) + (4 β3/π) +…..**

**Ans:Β **

**Method 1:**

The AP (4 β1/π) + (4 β2/π) + (4 β3/π) +…..Β can be re-arranged as:

(4 + 4 + 4 + 4 + …..Β up to π terms ) – (1 + 2 + 3+ …. upto π terms)

Therefore sum of n terms,

=

=

=

**Therefore, the sum of π term is = .**

**Method 2:**

The AP is given as:

Its first term,

Common difference,

We know that Sum of n terms of an AP is given by:

By substituting values of ‘a’ and ‘d’, we get:

**Therefore, the sum of π term is = .**

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