CAT 2004 — QA Question 27
Consider the sequence of numbers a1, a2, a3, ... to infinity where a1 = 81.33 and a2 = –19 and aj = aj–1 – aj–2 for j ≥ 3. What is the sum of the first 6002 terms of this sequence?
Answer & solution
Correct answer: 62.33
- A
-100.33
- B
-30.00
62.33
- D
119.33
a1 = 81.33
a2 = –19
a3 = a2 âââââââ- a1 = –100.33
a4 = a3 âââââââ- a2 âââââââ = –81.33
a5 = a4 âââââââ- a3 âââââââ = 19
a6 = a5 âââââââ- a4 âââââââ = 100.33
a7 = a2 âââââââ- a5 âââââââ = 81.33
a8 = a7 âââââââ- a6 = –19
We can 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 6001st and 6002nd terms = 81.33 – 19 = 62.33
Hence, option (c).