In an arithmetic series the 20th term is 14 and the 40th term is -6. Find the 10th term.
How do I do this?
$\displaystyle U_n = a+(n-1)d$
Where
- $\displaystyle U_n$ is the nth term
- $\displaystyle a$ is the first term
- $\displaystyle n$ is the number of terms
- $\displaystyle d$ is the common difference.
$\displaystyle U_{20} = a + (20-1)d = 14$ (eq1)
$\displaystyle U_{40} = a + (40-1)d = -6$ (eq2)
Solve simultaneously. It would be easiest to eliminate a first
$\displaystyle U_{20} > U_{40}$ which means $\displaystyle d<0$
I took equation 2 from equation 1:
$\displaystyle (a+19d) - (a+39d) = 14 - (-6)$
$\displaystyle -20d = 20 \: \therefore \: d = -1$
$\displaystyle a+19(-1) = 14 \: \therefore \: a = 33$
$\displaystyle U_{10} = a + 9d = 33-9 = 24$
By definition: $\displaystyle a = 5x$
$\displaystyle U_2 = 20 = a+d $
$\displaystyle U_3 = 3x = a+2d$
You can then solve U_2 and U_3 simultaneously. Eliminate d since we don't need to find it
I get $\displaystyle x=5$ and so $\displaystyle a = 25$, $\displaystyle U_2 = 20$ and $\displaystyle U_3 = 15$