Series-6 (CAT-2004)

Consider the sequence of numbers a₁, a₂, a₃,... to infinity, where a₁ = 81.33 and a₂ = -19 and a_j = a_j-1 – a_j-2. What is the sum of first 6002 terms of this sequence?
(1)-100.33                  (2)-30.00                    (3)62.33                     (4)119.33
Solution follows here:

Solution:

a_j = a_j-1 – a_j-2 => a₃ = a₂ – a₁;

a₄ = a₃ – a₂ = a₂ – a₁ – a₂ = -a₁;

a₅ = a₄ – a₃ = -a₁ – (a₂ – a₁) = -a₂;

a₆ = a₅ – a₄ = -a₂ – ( – a₁) = a₁-a₂;

a₇ = a₆ – a₅ = a₁-a₂ – ( –a₂) = a₁;

As a₇ is being equal to a₁, from a₇ onwards, the series is starting to repeat.

=> First six terms are getting repeated

a₁+a₂+a₃+a₄+a₅+a₆ = a₁+a₂+( a₂ – a₁)+( -a₁)+( -a₂)+( a₁-a₂) = 0

Sum of first six terms is zero => sum of next six terms is also zero…

So sum of first 6000 terms is also zero, as 6000 is a multiple of six.

Sum of first 6002 terms = (sum of first 6000 terms)+( sum of first 2 terms)

= 0+( a₁+a₂) = 81.33-19 = 62.33

Answer (3)