If a1, a2, a3, ..., an are in A.P. and a1 = 0, then the value of ( a3/a2 + a4/a3 +... + an/a(n-1)) - a2(1/a2 +1/a3 +...+ 1/a(n-2)) is

Question

If a₁, a₂, a₃, ..., an are in A.P. and a1 = 0, then the value of \(\big(\frac{a_3}{a_2} +\frac{a_4}{a_3} + ...+ \frac{a_n}{a_{n-1}}\big)\) - a₂ \(\big(\frac{1}{a_2} +\frac{1}{a_3} + ...+ \frac{1}{a_{n-2}}\big)\) is

(a)  n + \(\frac{1}{n}\) 

(b)  n + \(\frac{1}{n-1}\) 

(c)  (n - 1) + \(\frac{1}{n-1}\) 

(d)  (n - 2) + \(\frac{1}{n-2}\) 

Answer 1

Answer : (d)  (n - 2) + \(\frac{1}{n-2}\)  

Given, a₁, a₂, a₃, ..... , an are in A.P and a₁ = 0. 

Let d be the common difference of the the A.P. So, 

a₂ = a₁ + d = d 

∴ a₃ = a₁ + 2d = 2d, 

a₄ = 3d, a₅ = 4d, .... , a_n = (n – 1)d 

 ∴ \(\big(\frac{a_3}{a_2} +\frac{a_4}{a_3} + ...+ \frac{a_n}{a_{n-1}}\big)\) - a₂ \(\big(\frac{1}{a_2} +\frac{1}{a_3} + ...+ \frac{1}{a_{n-2}}\big)\) 

= \(\big[ \)\(\frac{2d}{d}\) + \(\frac{3d}{2d}\) +... + \(\frac{(n-1)d}{(n-2)d}\)\(\big]\) - d\(\big[ \)\(\frac{1}{d}\) + \(\frac{1}{2d}\) +...+ \(\frac{1}{(n-3)d}\)\(\big]\)

= 1 + 1 + 1 .... to (n – 2) term 

=  (n - 2) + \(\frac{1}{n-2}\)