f(x) = -2x + 4

Let S = F(1) + f(2) + ...+ f(10)

f(1) = -2*1 + 4 = 2

f(2) = -2*2+4 = 0

f(3) = -2*3 + 4 = -2

f(4) = -2*4 + 4 = -4

f(5) = -2*5 + 4 = -6

.......

We notice that S is an A.P such that a1= 2 n= 10 and r= -2

==> S= 2 + 0 + -2 + -4 + ....+ -16

= 2 + 0 + -2(1+2+....+8)

= 2 + -2*(n(n+1)/2)

= 2 + -2*(36)

= 2- 72 = -70

**==> S = -70**

To evaluate the sum, we'll have to substitute x by each of the values: 1,2,3,...,10.

f(1) = -2*1 + 4

f(2) = -2*2 + 4

f(3) = -2*3 + 4

......................

f(10) = -2*10 + 4

We'll calculate the sum:

f(1) + f(2) +..+ f(10) = -2*(1+2+...+10) + 10*4

We'll remove the brackets and we'll substitute the sum of the first 10 natural numbers, by the product:

1+2+...+10 = (1+10)*10/2

1+2+...+10 = 11*5

1+2+...+10 = 55

f(1) + f(2) +..+ f(10) = -2*55 + 40

f(1) + f(2) +..+ f(10) = -110 + 40

**f(1) + f(2) +..+ f(10) = -70**

f(x) = -2x+4

f(x+1) - f(x) = -2(x+1)+4-(-2x+4) = = -2x-2+4+2x-4 = -2 is the common difference.

Therfore this an AP.

Sn + {f(1)+f(n)}n/2

So ,S10 = {f(1)+f(10)}* 10/2

S10 = {-2*1+4 + (-2*10+4)}10/2

S10 = {2-16}10/2 = -14*10/2 = -70

S10 = -70.