Very easy!!!!!!!!
Take the first term to be a and the last term to be b. Let the common difference be d .
then, mulitiplying 1st and last you get: ab
and multiplying 2nd and 2nd last you get ab+d(b-a-d).
Now, if d>0, b>a and also b> a+d
so, b-a-d >0
and hence d(b-a-d) > 0
If d<0
b<a
and b<a+d,
so, b-a-d<0
so d(b-a-d)>0
so, product of 2nd and 2nd last = ab+some positive number and is hence greater than ab[product of 1st and last]
So,
ab < ab+d(b-a-d)
so, ab+d(b-a-d)< ab + 2d(b-a-d)
i.e. product of 2nd and 2nd last< product of 3rd and 3rd last.
IN this way the product of equidistant terms keep on incresing as one moves closer to the middle term.
please rate me..........