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.
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!