CAT 2004 | Question: 66

Question

Consider the sequence of the numbers $a_1, a_2, a_3, \dots$ to infinity where $a_1 = 81.33$ and $a_2 = -19$, and $a_j = a_{j-1} - a_{j-2}$ for $j \geq 3$. What is the sum of the first $6002$ terms of this sequence?

• A. $-100.33$
• B. $-30.00$
• C. $62.33$
• D. $119.33$

Answer 1

$a_{1}$= 81.33                $a_{2}$= –19

and $a_{ j }$ = $a_{ j-1 }$ - $a_{ j-2 }$  for j ≥ 3

$a_{3}$ = $a_{2}$ - $a_{1}$ = -19 - 81.33 = -100.33

Similarly,
$a_{4}$= $a_{3}$ - $a_{2}$ = -100.33 -(-19) = –81.33 

$a_{5}$ = 19

$a_{6}$ = 100.33

$a_{7}$ = 81.33

$a_{8}$ = –19

Here we can clearly see that the sequence repeats itself after every 6 terms.

Sum of the first 6 terms of the sequence = 0

Thus, the sum of the first 6000 terms of this sequence = 0

 

The sum of the $6001^{st}$ and $6002^{nd}$ terms = 81.33 – 19 = 62.33

 

Hence,Option(3)62.33 is the correct choice.