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If a1, a2, a3, ... an be an A.P. of non-zero terms, then find the sum: \(\frac{1}{a_1a_2}\) + \(\frac{1}{a_2a_3}\) + ... + \(\frac{1}{a_{n-1}a_n}\). |
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Answer» Let a2 – a1 = a3 – a2 = ........ = an – an – 1 = d (common difference) Then, \(\frac{1}{a_1a_2}\) + \(\frac{1}{a_2a_3}\) + ... + \(\frac{1}{a_{n-1}a_n}\) = \(\frac{1}{d}\)[\(\frac{d}{a_1a_2}\) + \(\frac{d}{a_2a_3}\) + ... + \(\frac{d}{a_{n-1}a_n}\)] = \(\frac{1}{d}\)[\(\frac{a_2-a_1}{a_1a_2}\) + \(\frac{a_3-a_2}{a_2a_3}\) + ... + \(\frac{a_n- a_{n-1}}{a_{n-1}a_n}\) ] = \(\frac{1}{d}\)[ \(\frac{1}{a_1} -\frac{1}{a_2}\) + \(\frac{1}{a_2} -\frac{1}{a_3}\)+ ... + \(\frac{1}{a_{n-1}} -\frac{1}{a_n}\)] = \(\frac{1}{d}\)[\(\frac{1}{a_1} -\frac{1}{a_n}\)] = \(\frac{1}{d}\)[ \(\frac{a_n-a_1}{a_1a_n}\)] = \(\frac{1}{d}\)[ \(\frac{(a_1 +(n-1)d) -a_1}{a_1a_n}\)] = \(\frac{n-1}{a_1a_n}\). |
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