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# Help!!!!!

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1.Part (a): Find the sum \(a + (a + 1) + (a + 2) + \dots + (a + n - 1)\)in terms of a and n. Part (b): Find all pairs of positive integers (a,n) such that \(n \ge 2\) and \([a + (a + 1) + (a + 2) + \dots + (a + n - 1) = 100.\)

2. Find x such that \(\lceil x \rceil + x = \dfrac{23}{7}\) . Express x as a common fraction.

Thanks!!! (give me a thumbs up!)

Sep 9, 2017
edited by MIRB16  Sep 9, 2017

#1
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Do u want to get cuaght?

Sep 9, 2017
#2
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2)

Solve for x over the real numbers:
x + abs(x) = 23/7

Hint: | Isolate terms with abs(x) to the left hand side.
Subtract x from both sides:
abs(x) = 23/7 - x

Hint: | Eliminate the absolute value.
Split the equation into two possible cases:
x = 23/7 - x or x = x - 23/7

Hint: | Look at the first equation: Isolate x to the left-hand side.
2 x = 23/7 or x = x - 23/7

Hint: | Solve for x.
Divide both sides by 2:
x = 23/14 or x = x - 23/7

Hint: | Look at the second equation: Isolate x to the left-hand side.
Subtract x from both sides:
x = 23/14 or 0 = -23/7

Hint: | Look for a false statement.
0 = -23/7 is trivially false:
x = 23/14

1)

I can only see two pairs of a, n that satisfy the given conditions:

9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100, where a = 9 and n = 8

18 + 19 + 20 + 21 + 22 = 100, where a = 18 and n = 5

Sep 9, 2017
#3
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2)

x + abs(x) = 23/7, solve for x

Split it into 2 equations as follows:

+x + x = 23/7.............(1)

- x + x = 23/7.............(2)

In equation (1) you have:

2x = 23/7  divide both sides by 2

x = (23/7) / 2

x = 23/7 x 1/2

x = 23/14

In equation (2) you have:

-x + x = 23/7

0 = 23/7 Discard this, since it is false.

Sep 9, 2017
#4
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Here's the second one

We have  that

ceiling (x)  +  x  =  23 / 7     .....where the "ceiling" function represents the smallest integer ≥ x

Note that if   x  =  9/7  we have that

ceiling (9/7)  +  9/7  =

ceiling ( 1 + 2/7)  +  9/7  =

2    +  9/7   =

14/7 + 9/7  =

23/7

Sep 10, 2017