Let an,nϵN is an A.P. with common difference 'd' and all whose terms are non-zero. If n approaches infinity, then the sum 1a1a2+1a2a3+...+1anan+1 will approach
A
1a1d
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B
2a1d
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C
12a1d
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D
a1d
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Solution
The correct option is C1a1d 1a1a2+1a2a3+... =1d(a2−a1a1a2+a3−a2a2a3+...) =1d(1a1−1a2+1a2−1a3+...) =1a1d