Add any of the 5 numbers from 2, 6, 10, 14, 18, 22, 26, 30, 34, 38 and get the sum as 100.

Question

aum · Answer

I am thinking it may not be possible.

aum · Answer

The numbers are in arithmetic sequence and there are 10 terms.
The kth term is: $2 + (k-1)d = 2 + (k - 1)4 = 2 + 4k - 4 = 4k - 2$
 where $1 \le k \le 10$

anonymous · Answer

k thank you

aum · Answer

Assume $k_1, k_2, k_3, k_4, k_5$ represents those k values whose terms add up to 100.
Then, 
$4(k_1+k_2+k_3+k_4+k_5) - 10 = 100 \
4(k_1+k_2+k_3+k_4+k_5)  = 110 \
(k_1+k_2+k_3+k_4+k_5) = 27.5
$
But $k_1, k_2, k_3, k_4, k_5$ are all integers from 1 to 10.  They cannot add up to 27.5.  So there is no solution.

aum · Answer

In case the third line above is not clear, since the $k^{th}$ term is $4k - 2$, the 
$k_1^{th}$ term is: $4k_1 - 2$
$k_2^{th}$ term is: $4k_2 - 2$
etc...

aum · Answer

I don't know if the same number can be picked more than once.

triciaal · Answer

i don't have a solution but what I had was the average is 100/5 = 20
the numbers chosen deviate as -18, -14,-10,-6,-2, 2, 6, 10,14, 18 then as @aum states there is an arithmetic progression with d = 4

aum · Answer

I think tricia's  method changes the problem to: pick 5 numbers from -18, -14,-10,-6,-2, 2, 6, 10,14, 18 that add up to zero.

triciaal · Answer

correction  d =-4

aum · Answer

I think it is d = +4.

triciaal · Answer

I need to go to bed  thanks add 4

anonymous · Answer

This is the first time I have attempted a "proof" on this site.  
Refer to the Mathematica calculation attached.