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Question

If a,a2,a3,....,an are first n terms of an AP with first term a=0 and common difference d0 Then value of a3a2a2+a4a2a3+a5a2a4....+ana2an1 is

A
n
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B
(n1)n2
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C
n2
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D
n(n+1)2
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Solution

The correct option is C n2
Given, a3a2a2+a4a2a3+a5a2a4+............+ana2an1
Since, an=a+(n1)d
a2=a+d,a3=a+2d,a4=a+3d.............. and an1=a+(n2)d
Substitute the values of a2,a3,a4,a5,............,an1,an in given statement. we get,
a+2d(a+d)a+d+a+3d(a+d)a+2d+a+4d(a+d)a+3d+............+a+(n1)d(a+d)a+(n2)d
da+d+2da+2d+3da+3d+............+(n2)da+(n2)d
given, a=0
dd+2d2d+3d3d+............+(n2)d(n2)d
1+1+1+........(n2)times=n2
Option C is correct.

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