Solve 2+4+6 +⋯ +48−3−5−7 − … −51

WITH EXPLANATION PLEASE!!!

-98

Step-by-step explanation:

Given:

2+4+6 +⋯ +48−3−5−7 − … −51

Find solution:

(2+4+6 +⋯ +48)−(3+5+7+ …+51)

Computation:

an = a + (n-1)d

For

(2+4+6 +⋯ +48)

48 = 2 + (n-1)2

n=22

For

(3+5+7+ …+51)

51 = 3 + (N-1)2

N = 24

Sn = n/2(a+an)

So

22/2(2+48)-24/2(3+51)

550 - 648

-98

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v