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Question

If a1,a2,a3...anare in AP and a1=0, then the value of a3a2+a4a3+.+anan-1-a21a2+1a2+....+1an-2 is equal to


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Solution

Finding the value of (a3a2+a4a3+.+anan-1)-a2(1a2+1a2+....+1an-2):

Given that a1,a2,a3...an are in AP

and a1=0

a2=a1+d=da3=a1+2d=2dan-1=(n-2)dan=(n-1)da3a2+a4a3+.+anan-1-a21a2+1a2+....+1an-2

=2dd+3d2d+...(n-1)d(n-2)d-d1d+12d+...+1(n-3)d

=2+32+...(n-1)(n-2)-(1+12+...+1(n-3)) ={(1+1)+(1+12)+...(1+1(n-2)}-{1+12+13+...+1(n-3)}

=1+1+1+....(n-2) Terms

+{1+12+13+....+1(n-3)+1(n-2)}-{1+12+13+...+1(n-3)}=(n-2)+1(n-2)

Hence, the value is (n-2)+1(n-2).


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