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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: 

 

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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3+0 Answers

 #1
avatar+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

 

 

cool cool cool

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

Thank you so much this was super helpful!

 
AnonymousConfusedGuy  Jan 10, 2018
 #2
avatar+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)     

 

 

cool cool cool                   

 
CPhill  Jan 10, 2018

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