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? (CAT 2004)
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? (CAT 2004)
The correct option is C 62.33
The terms of the given sequence are as follows :
a1=81.33 a7=a1
a2=−19 a8=a2
a3=a2−a1 a9=a3
a4=−a1 a10=a4=−a1
a5=−a2 a11=a4=−a2
a6=−a2+a1 a12=a6=−a3 and so on.
The sum of the first six terms, the next six terms and so on is 0.
The sum of the first 6002 terms can be written as the sum of first 6000 terms + 6001st term + 6002ndterm. From the above explanation, the sum of the first 6000 terms is zero, 6001 term will be a1 and 6002nd term will be a2. The sum of the first 6002 terms will be a1+a2=81.33+(−19)=62.33.
The terms of the given sequence are as follows :
a1=81.33 a7=a1
a2=−19 a8=a2
a3=a2−a1 a9=a3
a4=−a1 a10=a4=−a1
a5=−a2 a11=a4=−a2
a6=−a2+a1 a12=a6=−a3 and so on.
The sum of the first six terms, the next six terms and so on is 0.
The sum of the first 6002 terms can be written as the sum of first 6000 terms + 6001st term + 6002ndterm. From the above explanation, the sum of the first 6000 terms is zero, 6001 term will be a1 and 6002nd term will be a2. The sum of the first 6002 terms will be a1+a2=81.33+(−19)=62.33.
