1. ## arithmetic progression

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

2. 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 $\displaystyle a_n=a_1+(n-1)d$

Let $\displaystyle x=a_3$

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

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

$\displaystyle x=3+2d$

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

Do a similar calculation using:

Let $\displaystyle a_7=2x$

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

$\displaystyle 2x=3+6d$

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

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

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

4. 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.

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

Setting them equal to each other, we have

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

$\displaystyle x=6$

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

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

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

And there you have it!

5. 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

6. thanks everyone. much appreciated.