CAT Exam Model Paper 7 with solutions for free online practice

Question : 16 of 60

Marks: +1, -0

A sequence of non-negative integers is given such that

t_{1}=150

and

t_{n}=t_{n-2}-t_{n-1}

for n > 2. For what value of t., would the sequence have the maximum possible number of terms?

Validate

Solution:

All the terms of the sequence have to be non-negative integers As soon as we get a negative term it would mean that the sequence terminates at the previous term.
Let's write the first few terms:

\table t_{3}=150-t_2, t_{4}=2 t_{2}-150 ; t_{5}=300-3 t_{2} , t_{6}=5 t_{2}-450 ; t_7=750-8 t_2, t_{8}=13 t_{2}-1200 ; t_{9}=1950-21 t_{2}, t_{10}=34 t_{2}-3150

Now let's try to make as many of them positive as possible:

{150-t_{2}≥0} { { \or }} {150≥ t_{2}}

{2 t_{2}-150 ≥0} { { \or }} {t_{2} ≥ 75}

{300-3 t_{2} ≥ 0} { { \or } 100≥ t_{2}}

5 t_{2}-450 ≥ 0

t_{2} ≥ 90

750-8 t_{2} ≥ 0

93.75 ≥t_{2}

13 t_{2}-1200 ≥ 0

t_{2} ≥ 92.30

t_2

must be greater than 92 and less than 94, for the first 8 terms to be positive
So when

t_2

= 93. the sequence would have exactly 8 terms.
For every other value of

t_2

, the number of terms would be less than 8.
So the answer is 93.

Go to Question:

Prev Question

Next Question