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# Hey! Really sorry, but just in case this has slid to the bottom and nobody sees it, I kindof need some urgent help with this:

0
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+378

Hey!  Really sorry, but just in case this has slid to the bottom and nobody sees it, I kindof need some urgent help with this:

https://web2.0calc.com/questions/part-a-find-the-sum

If anyone could help that would be the best!

AnonymousConfusedGuy  Jan 10, 2018
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#1
+80793
+1

Part (a): Find the sum

a + (a + 1) + (a + 2) + ... + (a + n - 1)

in terms of a and n.

We can write

( n ) a   +   ( 1 + 2 + 3 + ... +  (n - 1) )

The sum of the second part is

(n - 1) ( n)  / 2

Putting these together, we have that the sum  is :

(n)a  +  (n) ( n-1) / 2    =

(n)   [   a  +  (n - 1) / 2  ]  =

(n) ( 2a  +  n - 1)  /  2

CPhill  Jan 10, 2018
edited by CPhill  Jan 10, 2018
#3
+378
+1

Thank you so much this was super helpful!

AnonymousConfusedGuy  Jan 10, 2018
#2
+80793
+1

Part b

Using  the result from the first part, we have that

(n) ( 2a  +  n - 1)  /  2  =  100     multiply both sides by 2

(n) (2a + n - 1)  =  200

Factors  of 200  are      1 | 2 | 4 | 5 | 8 | 10 | 20 | 25 | 40 | 50 | 100 | 200

So we have the following possibilities for

n               (2a +  n  - 1)     =       200

2                  100                   a  is not an integer  if n  = 2

4                    50                   a is not an integer if n  = 4

5                    40                   a  =  18

8                    25                   a  =  9

10                  20                   a is not an integer if n  = 10

We can stop here....all other possible factor combinations for 200 result  in negative values for a

So....  ( a , n)    =   ( 18, 5)   and ( 9 , 8)

CPhill  Jan 10, 2018

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