arithmetic progression

**finnkeers** · May 18th 2010, 11:40 AM

how do i find common difference if the first term is 3 and the seventh term is twice the third term. bit of help.

**masters** · May 18th 2010, 11:52 AM

Originally Posted by finnkeers

how do i find common difference if the first term is 3 and the seventh term is twice the third term. bit of help.

Hi finnkeers,

Use $a_n=a_1+(n-1)d$

Let $x=a_3$

$a_3=a_1+(n-1)d$

$x=3+(3-1)d$

$x=3+2d$

${\color{red}d=\frac{x-3}{2}}$

Do a similar calculation using:

Let $a_7=2x$

$a_7=3+(7-1)d$

$2x=3+6d$

${\color{red}d=\frac{2x-3}{6}}$

Now set the two d's equal to each other and solve.

**finnkeers** · May 18th 2010, 12:33 PM

sorry but now i have two variables so how do i solve for d by getting rid of x

**masters** · May 19th 2010, 04:02 AM

Originally Posted by finnkeers

sorry but now i have two variables so how do i solve for d by getting rid of x

Well, let's see.

$d=\frac{x-3}{2}$ and $d=\frac{2x-3}{6}$

Setting them equal to each other, we have

$\frac{x-3}{2}=\frac{2x-3}{6}$

$x=6$

If x = 6, then substituting into either equation with d, we get

$d = \frac{6-3}{2} =\boxed{\frac{3}{2}}$

$d = \frac{2(6)-2}{6}=\frac{9}{6}=\boxed{\frac{3}{2}}$

And there you have it!

**BobBali** · May 19th 2010, 05:31 AM

first term (u1) = 3

u1 + 6d = 7th term But 7th term = 2(u1+2d)

set equations equal to one another:

3 + 6d = 2(3+2d)
3+6d = 6+4d
6d-4d = 6-3
2d= 3

Therefore, d=3/2

**finnkeers** · May 19th 2010, 06:56 AM

thanks everyone. much appreciated.

Thread: arithmetic progression

LinkBack

Thread Tools

Display

arithmetic progression

Similar Math Help Forum Discussions

Arithmetic Progression or Arithmetic Series Problem

Arithmetic progression help..

Arithmetic Progression & Geometric Progression

arithmetic progression--need help 2

Arithmetic Progression X.x

Search Tags