How many terms of the A.P $ 17,15,13,..................... $ must be added to get the sum 72? Explain the double answer.

Hint: In this we can use the sum formula of A.P to find the value of the required number of terms.If first term of A.P is a , common difference is d and number of terms is n then sum(S) of A.P is
 $ S=\dfrac{n}{2}\left\{ 2a+(n-1)d \right\} $

Complete step-by-step answer:
Given A.P is $ 17,15,13,..................... $
First term is 17
Common difference is(d) = 15-17 = 13-15 = -2
Let the number of terms is n to get the sum of 72.
Hence we can write
 $ \Rightarrow S=\dfrac{n}{2}\left\{ 2a+(n-1)d \right\} $
 $ \Rightarrow 72=\dfrac{n}{2}\left\{ 2\times 17+(n-1)\times -2 \right\} $
 $ \Rightarrow 144=n\left\{ 34-2n+2 \right\} $
 $ \Rightarrow 144=n\left\{ 36-2n \right\} $
 $ \Rightarrow 144=36n-2{{n}^{2}} $
 $ \Rightarrow 2{{n}^{2}}-36n+144=0 $
 $ \Rightarrow 2({{n}^{2}}-18n+72)=0 $
 $ \Rightarrow {{n}^{2}}-18n+72=0 $
Now we can factorize it by splitting the middle term.
So we can split 18 in two terms whose sum equal to 18 and product equal to 72.
So we can write it as
 $ \Rightarrow {{n}^{2}}-12n-6n+72=0 $
 $ \Rightarrow n(n-12)-6(n-12)=0 $
 $ \Rightarrow (n-12)(n-6)=0 $
So n = 12 or n= 6.
If we write A.P upto 6 terms then sum is
 $ S=17+15+13+11+9+7=72 $
If we write A.P upto 12 terms then sum is
 $ S=17+15+13+11+9+7+5+3+1+(-1)+(-3)+(-5)=72 $
So we can see that the sum of 6 terms and sum of 12 terms is 72. Because it is decreasing A.P. So when we write it upto 12 terms we can see the last 6 terms cancel each other.
So both answers are valid.

Note:In general Arithmetic series is a series in which the difference between two consecutive terms is the same. In the given question the difference between two consecutive terms is negative. So it is decreasing A.P.