Try hard, but not too hard!

2 is the 27th term of the AP. 4 is the 174th term or the 27th term from the back of the AP.

The sum of mirror image terms of an AP (sum of $n^{th}$ term from the front and $n^{th}$ term from the back of the AP) is always equal to the sum of the first and last terms.

So sum of AP = $(first term + last term)\times (number of terms)/2$

= $(27^{th} term+27^{th}lastterm)\times 200/2$

= $(2+4)\times 200/2$

= $600$

Given: $n=200$ , $a_{27}=2$ and $a_{174}=4$

solving for the common difference, d:

$a_n=a_m+(n-m)d$ $\large \implies$ $a_{174}=a_{27}+(174-27)d$ $\large \implies$ $4=2+147d$ $\large \implies$ $d=\dfrac{2}{147}$

solving for a_1:

$a_{27}=a_1+(27-1)d$ $\large \implies$ $2=a_1+26 \left(\dfrac{2}{147}\right)$ $\large \implies$ $a_1=\dfrac{242}{147}$

solving for the sum:

$s=\dfrac{n}{2}[2a_1+(n-1)d]$ $\large \implies$ $s=\dfrac{200}{2} \left(2 \left(\dfrac{242}{147} \right) +199 \left(\dfrac{2}{147} \right) \right)=$ $\boxed{600}$

Given:

$a+26d=2$ .... $(1)$
$a+173d=4$ .... $(2)$

Adding $(1)$ and $(2)$ .

$a+26d+a+173d=2a+199d=6$

Sum of 200 terms:

$S_{200}=100(2a+199d)=100×6=\boxed{600}$

$s_{200} = \frac{a_1 + a_{200}}{2} \cdot 200 = \frac{a_{27} + a_{174}}{2} \cdot 200 =\frac{2+ 4}{2} \cdot 200 = 600$

$n=200$

$a_{27}=2$

$a_{174}=4$

$4=2+(174-27)d$

$d=$ $\frac{2}{147}$

$a_{200}=4+(200-174)($ $\frac{2}{147})$ $=$ $\frac{640}{147}$

$4=$ $a_1+(174-1)($ $\frac{2}{147})$

$a_1=$ $\frac{242}{147}$

$S=\frac{n}{2}(a_1+a_n)$ $=($ $\frac{200}{2})$ $(\frac{242}{147}+\frac{640}{147}$ ) $=600$

$(1) 2 = A + 27D \\ (2) 4 = A + 174 \\ (2)-(1) 2 = (174-27D) = 147D \\ D = \frac{2}{147} \\ 147 = 3*7*7 \\ A_{(1)} = 2-\frac{27*2}{147} = 2 - \frac{3*3*3*2}{3*7*7} = 2 - \frac{3*3*2}{7*7} = 2-\frac{18}{49} = \frac{80}{49} \\ A_{(2)} = 4 -\frac{174*2}{147} = 4 - \frac{2*3*29*2}{3*7*7} = 4 - \frac{2*29*2}{7*7} = 4-\frac{116}{49} = \frac{80}{49}\\ A = \frac{80}{49} \\ \\ 200A + (200)(201)(0.5)D = X \\ X = 200A + 20100D = 200* \frac{80}{49} + 20100*\frac{2}{147} = \boxed{600}$

Note that an arithmetic series is equal to the average of the first and last terms multiplied by the number of terms. Since the difference of the 27th term from the first term is the same as the 200th term from the 174th, the average of the first and last terms is $\frac{2+4}{2}=3$ . This means that the answer is $3\cdot200=\boxed{600}$ .