Sequence and series - Arithmetic mean question have been breaking my head

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SUMMARY

The problem involves finding two consecutive numbers removed from the sequence 1 to n, such that the arithmetic mean (A.M) of the remaining numbers equals 105/4. The solution reveals that n equals 50, and the removed numbers are 7 and 8. The approach utilizes the formula for the A.M of a sequence and requires solving a quadratic equation derived from the conditions provided.

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Homework Statement


Two consecutive numbers from 1,2,3...n are removed A.M of remaining numbers is 105/4. Find n and those numbers removed .

Homework Equations



Answer

n = 50
those numbers are 7 and 8

The Attempt at a Solution



I solved this question like a few weeks ago but now it escaped my brain I have no idea how .. and I am so annoyed now you know how it feels right?
Ok so here's what I got
The A.M of the total numbers would be
n(n+1) / 2 / n

that gives (n+1)/2

now what we get after removing the numbers
let the numbers be x and x+1
therefore their sum = 2x+1

now A.M after removing numbers

n(n+1)/2 - (2x+1) / (n-2) = 105/4

heres one I am not able to figyre out another relation PLease help somebody
 
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Multiply both sides by 2(n-2).
\frac{\frac{n(n+1)}{2} - (2x+1)}{n - 2} \cdot 2(n - 2) = \frac{105}{4} \cdot 2(n - 2)
Rearrange and you'll find yourself with a quadratic in n. Pretend for a moment that x is a constant, and solve for n using the quadratic formula. Then you'll need to guess-and-check some values of x to make the discriminant a perfect square.
 

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