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If the terms a1, a2,..., a(n+1) form an arithmetic progression, then the value of 1/(a1a2) + 1/(a2a3) +... + 1/(aₙ a(n+1)) is

1. (n−1)/(a1a(n+1))
2. 1/(a1a(n+1))
3. (n+1)/(a1a(n+1))
4. n/(a1a(n+1))

Correct answer: n/(a1a(n+1))

Solution

With common difference d, 1/(a_i a_{i+1}) = (1/d)(1/a_i - 1/a_{i+1}). Summing telescopes to (1/d)(1/a1 - 1/a_{n+1}) = (a_{n+1}-a1)/(d a1 a_{n+1}) = n/(a1 a_{n+1}).

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