If a1, a2, a3, ... an be an A.P. of non-zero terms, then find the sum: 1 /a1a2 + 1/a2a3 + ... + 1/a(n-1) an.

Question

If a₁, a₂, a₃, ... a_n be an A.P. of non-zero terms, then find the sum: \(\frac{1}{a_1a_2}\) + \(\frac{1}{a_2a_3}\) + ... + \(\frac{1}{a_{n-1}a_n}\).

Answer 1

Let a₂ – a₁ = a₃ – a₂ = ........ = a_n – a_{n – 1} = d (common difference)

Then,

  \(\frac{1}{a_1a_2}\) + \(\frac{1}{a_2a_3}\) + ... + \(\frac{1}{a_{n-1}a_n}\) = \(\frac{1}{d}\)[\(\frac{d}{a_1a_2}\) + \(\frac{d}{a_2a_3}\) + ... + \(\frac{d}{a_{n-1}a_n}\)]

= \(\frac{1}{d}\)[\(\frac{a_2-a_1}{a_1a_2}\) + \(\frac{a_3-a_2}{a_2a_3}\) + ... + \(\frac{a_n- a_{n-1}}{a_{n-1}a_n}\) ] = \(\frac{1}{d}\)[ \(\frac{1}{a_1} -\frac{1}{a_2}\) + \(\frac{1}{a_2} -\frac{1}{a_3}\)+ ... + \(\frac{1}{a_{n-1}} -\frac{1}{a_n}\)]

= \(\frac{1}{d}\)[\(\frac{1}{a_1} -\frac{1}{a_n}\)] =  \(\frac{1}{d}\)[ \(\frac{a_n-a_1}{a_1a_n}\)] 

= \(\frac{1}{d}\)[ \(\frac{(a_1 +(n-1)d) -a_1}{a_1a_n}\)] = \(\frac{n-1}{a_1a_n}\).