# Thread: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1thenx1=?

1. ## there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1thenx1=?

More precisely,
Let x1,x2,....x100 be positive nos. such that the sum of two adjacent nos. is k(constant). (xi +x (i+1) = k for all i)
If x10=1
x1=?

I solved it and my answer is one. But at the back it is showing k-1.

2. ## Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

Originally Posted by AaPa
More precisely,
Let x1,x2,....x100 be positive nos. such that the sum of two adjacent nos. is k(constant). (xi +x (i+1) = k for all i)
If x10=1
x1=?

I solved it and my answer is one. But at the back it is showing k-1.
$x_{2n}=1$

$x_{2n+1}=k-1$

1 is odd so $x_1=k-1$

3. ## Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

nice method. Thanks. I had done it with another method but wrongly. I corrected it-right answer.

4. ## Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

Hi,
When I was teaching I would discuss the sequence described in the attachment in many different classes. The main purpose was to show there is nothing mysterious about compound interest, examples 2 and 3.

cool