CAT 2004 — QA Question 27
Answer the following question based on the information given below.
In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks.
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
- 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).