# Arithmetic sequences, just a lil more help needed :)

• Feb 18th 2009, 06:35 AM
alyson
Arithmetic sequences, just a lil more help needed :)
hello, i have this maths question that i feel like i should be able to do easily but i cant
its arithmetic progressions (again), and the question is to find the sum of the first 8 digits, where the first two are the sum of 18, and the first four are the sum of 52
I'm not asking for all the work to be done, just the method needed so i can follow it and get the answer, which btw is 168 as it says in the back of my book, many thanks :)
• Feb 18th 2009, 06:55 AM
Amanda H
Using the information you have and the formula for the sum to n terms of an aritmetic sequence construct 2 equations with the variables a and d as follows:

First using sum of first 2 numbers is 18:
$\displaystyle S_{n} = \frac{n}{2}[2a + (n - 1)d]$
$\displaystyle 18 = \frac{2}{2}[2a + (2 - 1)d]$
$\displaystyle 18 = 2a +d$

Then the second piece of information sum to 4 is 52:
$\displaystyle 52 = \frac{4}{2}[2a + (4 - 1)d]$
$\displaystyle 52 = 2[2a + 3d]$
$\displaystyle 52 = 4a + 6d$

All you need to do now is solve simultaneously to find a and d. Then using these values use the same formula again to find the sum to 8 terms.
• Feb 18th 2009, 07:00 AM
Quote:

Originally Posted by alyson
hello, i have this maths question that i feel like i should be able to do easily but i cant
its arithmetic progressions (again), and the question is to find the sum of the first 8 digits, where the first two are the sum of 18, and the first four are the sum of 52
I'm not asking for all the work to be done, just the method needed so i can follow it and get the answer, which btw is 168 as it says in the back of my book, many thanks :)

Well , the first four terms of the progression are :

a , a+d , a+2d , a+3d

The sum of the first 2 terms :

a+a+d=18 , 2a+d=18 ------ 1

The sum of the first 4 terms :

a+a+d+a+2d+a+3d =52 , 4a+6d=52 ------ 2

Solve the system and you should get a=7 and d=4

$\displaystyle S_8=\frac{8}{2}(14+28)$
• Feb 18th 2009, 07:06 AM