Very Average
If each number in this list is the average of the two numbers before it, what is the value of a?
Problem
In the list below, each number (from $b$ onwards) is the mean of the two numbers before it.
$40$, $a$, $b$, $c$, $d$, $150$
Find the value of $a$
Student Solutions
Answer: 200
Using a number line
The average of two numbers is half way between them
Image
Image
Image
Label the shortest distance $n$
Image
Image
Image
Image
$11n=$ distance from $40$ to $150$
$11n=110\Rightarrow n=10$
$a=150+n+4n=200$
Using weights
$b$ is made from $a$ and $40$, weighted equally
Image
$c$ is made from $a$ and ($a$ and $40$), but $a$ should have weight equal to the weight of ($a$ and $40$). That is the same as using $a$ twice and ($a$ and $40$) once.
Image
$d$ is made from ($a$ and $40$) and (three $a$s and $40$), but again ($a$ and $40$) must be stretched to give equal weights:
Image
And the same for $150$:
Image
$150$ is the average of eleven $a$s and five $40$s
$150=\dfrac {11a+200}{16}$
$\Rightarrow 16\times150 = 11a+200$
$\Rightarrow 2400 = 11a+200$
$\Rightarrow 2200 = 11a$ so $a=200$
Using algebra starting from the beginning (weights algebraically)
$\dfrac{40+a}2=b$ and $\dfrac{a+b}2 = c$
So $\dfrac{a+\frac{40+a}2}2=c \Rightarrow \dfrac{40+3a}4=c$
$d=\dfrac{b+c}2 = \dfrac{\frac{40+a}2+\frac{40+3a}4}2 = \dfrac{2(40+a)+40+3a}8 = \dfrac {120 + 5a} 8$
$150=\dfrac{c+d}2 = \dfrac{\frac{40+3a}4+\frac {120 + 5a} 8}2 = \dfrac{80+6a + 120+5a}{16} =\dfrac{200+11a}{16}$
$\Rightarrow 16\times150 = 200 + 11a$
$\Rightarrow 2400 = 200 + 11a$
$\Rightarrow 2200 = 11a$ so $a=200$
Using algebra starting from the end
$150$ is the mean of $c$ and $d$ so $c +d =300\Rightarrow d = 300 - c$ (make $d$ the subject to get rid of $d$, then get rid of $c$ etc until only $a$ remains)
$d$ is the mean of $b$ and $c$ so $b+c=2d=2(300-c)$
$\Rightarrow b+c=600-2c \Rightarrow 3c=600-b \Rightarrow c = \frac13(600-b)$
$c$ is the mean of $a$ and $b$ so $a+b=2c=2\times \frac13(600-b)$
$\Rightarrow 3a+3b=1200-2b \Rightarrow 5b = 1200-3a \Rightarrow b = \frac15(1200-3a)$
$b$ is the mean of $40$ and $a$ so $40+a=2b=2\times\frac15(1200-3a)$
$\Rightarrow 200+5a = 2400-6a \Rightarrow 11a = 2200 \Rightarrow a=200$