Question 34

For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of 7th and 6th terms of this sequence is 517, what is the 10th term of this sequence?

Solution

It is given that in a Fibonacci sequence, from the third term on wards, each term in the sequence is the sum of the previous two terms in that sequence.

Let x and y be the 1st and 2nd term respectively.

3rd term = x+y

4th term = x+2y

5th term = 2x+3y

6th term = 3x+5y

7th term = 5x+8y

We know that difference of the squares of 6th and 7th terms is 517 = 47*11 .

And $$a^2-b^2=(a+b)(a-b)$$.

Applying above formula we get (8x+13y)(2x+3y) = 47*11.

So only possible way is (8x+13y)=47 and

2x+3y=11 .

Solving we get x=1 and y=3 .

Using the concept that every term is the sum of the previous two terms, as used in the beginning of the solution, we get 10th term as 21x+34y, which gives 10th term as 123.

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