if a1,a2....a2n form a decreasing A.P then
a12-a22+a32-a42...............-a2n2=
a)n(a12-a2n2)/2n-1
b)2n(a12-a2n2)/2n-1
c)2n(a12-an2)/n-1
d)n(a12-a2n2)/n+1
1a1,a2,...an are in decreasing AP. It means the common difference is -ve.
Let the common difference be -d.
a1 - a2 = a3-a4 = ... = d
a12-a22 + a32 - a42 + ... -a2n2
= (a1 - a2)(a1+a2) + ... + (a2n-1-a2n)(a2n-1+a2n)
= d(a1+a2+...a2n)
We know that sum of equidistant terms from both ends of an AP is same.
a1+a2n
= a2+a2n-1=...so on
d(a1+a2+...a2n) = nd(a1+a2n) ---1.
a2n = a1 + (2n-1)(-d) d = (a1-a2n)/2n-1
Putting this in 1, we get the required answer, which is A.
Hope this helps
ALL THE BEST
11SHORTCUT : (to save time)
Let a1,a2,...,an ≡ 3,2,1,0,-1,-2
2n=6 \Rightarrow n=3
Exp = 9 - 4 + 1 - 0 + 1 - 4 = 3 \leftarrow
a) 3(9-4)/6-1 = 15/5 = 3 \leftarrow
b) 6(9-4)/6-1 = 30/5 = 6
c) 6(9-4)/3-1 = 30/2 = 15
d) 3(9-4)/3+1 = 15/4 = 15/4
Hence a is correct
