Term no:       1,   2,   3,    4,   5,   6,  7,    8, ...

The series:  __, __, __, __, __, __, __, __, ...

.                  /_18_/

.                  /____52____/

From trial and error,

the 1st term = 7

the 2nd term = 18-7=11

common difference is 11-7=4

(Counter check by adding 4 consecutively to get 4th and 5th term to be 15 and 19 respectively.  It is confirmed that sum of first 4 terms is 52)

8th term = 7+ 4.(8-1) = 7+ 28 = 35

Sum of first 8 terms = 8/2 . (7 + 35) = 4 (42) = 168

Let a1, a2, a3, and a4 be the first four terms in an A.P.

Let ( r) be the common difference between terms.

Then :

a1=a1

a2= a1 + r

a3 = a1+ 2r

a4= a1 + 3r

Then given that :

a1 +a2 = 18

==> a1 + a1+ r = 18

==> 2a1 + r= 18 ..............(1)

Also, given that:

a1 + a2 + a3 + a4 = 52

==> a1 + (a1+ r)+ (a1+ 2r) + (a1+ 3r)= 52

==> 4a1 + 6r= 52

Divide by 2:

==> 2a1 + 3r = 26 ..............(2)

No we will solve the system using the elemintion method:

Subtract (1) from(2):

 ==> 2r = 8

==> r= 4

== > 2a1+ r = 18

==> 2a1= 18 - r

 ==> 2a1 = 18 - 4

=> 2a1= 14

==> a1= 7

==> a2= 7+ 4 = 11

==>a3 = 7 + 2*4 = 15

==> a4= 7 + 3*4 = 19

=> a5= 7+ 4*4 =  23

==>a6 = 7 +5*4 = 27

==> a7 = 7+ 6*4 = 31

==>a8 = 7+7*4 = 35

==> S8 = S4 + a5 + a6+ a7 + a8

                = 52 +23+27 + 31+ 35

                  = 168

Then,the sum of the first 8 terms is 168