C1 Tricky Questions Watch

Srathmore
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#1
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#1
The sum of the integers from 1 to n-1 is equal to the sum of integers from n+1 to 49. Find n.

If a and b are the first and last terms of an arithmetic series of r+2 terms, find the second and the (r+1)th term.

Ooo missed a question:

Given that y=ax^2 + bx - a^2 has gradient -4 at (-2,-13) find possible values for a and b!

Thanks xx
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Kolya
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#2
(1) \huge n=35
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*LEGEND*
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if you use the arithmetic sum formula : Sum = n/2 (First term + last term)

Then you get n^2 / 2 = (n^2 + 50n) / 2

But then im screwed cos the n^2's cancel and leave you with 50n = 0

hmmm
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footballhead
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#4
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how do u do it then
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Kolya
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#5
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\huge \frac{(n-1)(n)}{2} = \frac{(n+50)(49-n)}{2}
\huge (n-1)(n) = (n+50)(49-n)
\huge n^2-n = -n^2-n+2450
\huge 2n^2-n = -n+2450
\huge 2n^2 = 2450
\huge n^2 = 1225
\huge n = \pm{35}

\huge n = 35
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Codefusion
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#6
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remember there are two formulas to to get the sum one is n/2(a + L)
, we have the last so we are better off using this one.
first we have one arithmatic series on the LHS :

from 1 to n-1 which is the n factor becomes n-2
so n = n - 2
a = 1 (since it's 1 to n-1)
L = n-1

(n-2)/2 ( 1 + n - 1) = (n-2)/2 (n) = (n^2 - 2n)/2
on the RHS: another arithmatic series
from n+1 to 49
n = 49 - n - 1
a = n + 1
L = 49
(48 - n) / 2 ( n + 1 + 49 ) = (48 - n)/2 ( n + 50)
= (48n - n^2 + 2400 - 50n) / 2
= (2400 - n^2 - 2n)/2 (equate both sides)
(n^2 - 2n)/2 = (2400 - n^2 - 2n)/2
n^2 - 2n = 2400 - n^2 -2n
2n^2 = 2400
n^2 = 1200
n = sqr(1200) = 34.64 = 35
hope this helps.
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