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Question

Find the value of a+a+nda+a+d+a+a+nda+d+a+2d++a+a+nda+(n1)d+a+nd

A
0
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B
d
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C
n
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D
(n1)d
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Solution

The correct option is D n
Given a+a+nda+a+d+a+a+nda+d+a+2d+...+a+a+nda+(n1)d+a+nd

Raltionalize the denominators

=(a+a+nd)(1a+a+d+1a+d+a+2d+...+1a+(n1)d+a+2d)

=(a+a+nd)(1a+a+d×a+daa+da+1a+d+a+2d×a+2da+da+2da+d

+...+1a+(n1)d+a+nd×a+nda+(n1)da+nda+(n1)d)

=(a+a+nd)(a+daa+da+a+2da+da+2d(a+d)+a+3da+2da+3d(a+2d)

+...+a+nda+(n1)da+nd(a+(n1)d))

=(a+a+nd)(a+dad+a+2da+dd+a+3da+2dd

+...+a+nda+(n1)dd)

=(a+a+nd)(a+da+a+2da+d+a+3da+2dd

+....+a+nda+(n1)dd)

=(a+a+nd)(a+nda)d

=(a+nda)d=n

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