# The sum of the first two terms of an arithmetic progression is 18 and the sum of the first four terms is 52.  Find the sum of the first eight terms.

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

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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

neela | High School Teacher | (Level 3) Valedictorian

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The nth term of an AP is given by an = a1+n(-1)d, where a1 is the first term and d is the common difference.

Therefore sum of the first two terms of an AP = a1 + a1+d which is given to be18.

Or 2a1 +d = 18...(1).

The firest 4 terms of the AP = a1+a1+d+a1+2d+a1+3d  which is given to be 52.

So 4a1+6d = 52..(2).

Therefore eq (2) - eq 2(1) gives:

4a1+6d-2(2a1+d) = 52-2*18 = 16.

6d1-2d1 = 16.

4d1 = 16.

d1 = 16/4 = 4.

Therefore from (1) , 2a1+d1 = 18, or

2a1+4 = 18.

2a1 = 18-4 = 14

a1 = 14/2 = 7.

So a1 = 7.

We we know the sum Sn of the 1st n terms of an AP is given by:

Sn = {2a1+(n-1)d}n/2.

Therefore the sum of the first 8 terms  of the AP  is S8 = {2a1 + (8-1)d}8/2.

S8 = {2*7+(8-1)4}8/4 = {14+7*4}*4 = 42*4 = 168.

Therefore the sum of the 1st 8 terms of the given AP is S8=168.