# arithmetic series help

• Dec 4th 2008, 12:06 PM
me2612
arithmetic series help
This is the question:

The first term of an arithmetic series is 9 and the seventh term is three times the second term.
a) find the sum of the first 32 terms
b) how many more terms must be added for the sum of the A.P. to be 5040?

I have managed to work out d=6 and i think a) is 3264 but i have no idea what b) means PLEASE help !!!(Worried)

• Dec 4th 2008, 12:44 PM
Soroban
Hello, me2612!

Quote:

]The first term of an arithmetic series is 9
and the seventh term is three times the second term.

a) Find the sum of the first 32 terms.

We are told: . $a_1 = 9\:\text{ and }\:a_7 \:=\:3\cdot a_2$

In general: . $a_n \:=\:a_1 + (n-1)d$

We have: . $a_2 \:=\:a_1 + d \:\text{ and }\: a_7 \:=\:9 + 6d$

Then: . $9+6d \;=\;3(9 + d) \quad\Rightarrow\quad 3d \:=\:18 \quad\Rightarrow\quad \boxed{d \:=\:6}$

Formula: . $S_n \:=\:\tfrac{n}{2}\bigg[2a_1 + (n-1)d\bigg]$

Therefore: . $S_{32}\;=\;\tfrac{32}{2}\bigg[2(9) + 31(6)\bigg] \;=\;3264$

Quote:

b) How many more terms must be added for the sum of the A.P. to be 5040?
The sum of $k$ terms is 5040:

. . $\frac{k}{2}\bigg[2(9) + (k-1)(6)\bigg] \:=\: 5040 \quad\Rightarrow\quad k^2 + 2k - 1680 \:=\:0$

This factors: . $(k-40)(k+42) \:=\:0$

. . and has the positive root: . $k \,=\,40$

The sum of the first 32 terms is 3264.
The sum of the first 40 terms is 5040.

Therefore: . $40 - 32 \:=\:8$ more terms must be added.