If x is a positive integer, what is the sum of all the values of x for which 463 appears in the sequence

In a sequence of 10 terms, the first term is 1, the second term is x, and each term after the second is the sum of the previous two terms. For example, if x=11, the sequence would be 1, 11, 23, 35, 58, 93, 151, 244, 395. For some values of x, the number 463 appears in the sequence. If x is a positive integer, what is the sum of all the values of x for which 463 appears in the sequence?

the options are:
A-1156
B-1296
C-1248
D-1528
E-1283

thank you for your help

Since each term after the second is the sum of the two previous terms, we have that:
(1) a₁ = 1
(2) a₂ = x
(3) a₃ = 1 + x
(4) a₄ = 1 + 2x
(5) a₅ = 2 + 3x
(6) a₆ = 3 + 5x
(7) a₇ = 5 + 8x
(8) a₈ = 8 + 13x
(9) a₉ = 13 + 21x
(10) a₁₀ = 21 + 34x.

Then, setting each of these equal to 463 yields:
(2) x = 463
(3) 1 + x = 463 ==> x = 462
(4) 1 + 2x = 463 ==> x = 231
(5) 2 + 3x = 463 ==> x = 461/3
(6) 3 + 5x = 463 ==> x = 92
(7) 5 + 8x = 463 ==> x = 229/4
(8) 8 + 13x = 463 ==> x = 35
(9) 13 + 21x = 463 ==> x = 150/7
(10) 21 + 34x = 463 ==> x = 13.

Since we want integral values of x, we see that:
x = 463, 462, 231, 92, 35, and 13.

Therefore, the sum of the values of integral values of x:
463 + 462 + 231 + 92 + 35 + 13 = 1296.

I hope this helps!

The ten terms are:
1
x
x+1
2x+1
3x+2
5x+3
8x+5
13x+8
21x+13
34x+21

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Then we can run through the cases:
x = 463

x+1 = 463
=> x = 462

2x+1 = 463
=> x = 231

3x+2 = 463
=> x isn't an integer

5x+3 = 463
=> x = 92

8x+5 = 463
=> x isn't an integer

13x+8 = 463
=> x = 455/13 = 35

21x+13 = 463
=> x isn't an integer

34x+21 = 463
=> x = 13

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Finally,
463 + 462 + 231 + 92 + 35 + 13 = 1296

so it's (B).