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